6.12: Factors Impacting Organizational Design
- Identify aspects of the external environment that influence the design of an organization’s structure.
- Identify aspects of the internal environment that influence the design of an organization’s structure.
- Explain how business growth cycle affects organizational choices.
External Environment
The first factor to influence an organization’s design will be that of the external environment. The external environment consists of everything outside of organizations that can affect their performance and outcomes. Availability and need for raw materials, human resources, and financial resources are elements of the environment. Further key elements include customers and suppliers, competitors, cultural factors, and the types of regulatory frameworks or governmental influences on the organization.
The greater the number of external forces, the greater the complexity of the external environment. A pharmaceutical company operates in a very complex environment dealing with many groups such as doctors, hospitals, pharmacists, external research establishments, regulators, health insurance providers, labor markets, and many suppliers. In addition, a pharmaceutical company would deal with these groups in many countries with varying local economic conditions, health care arrangements, and regulatory regimes.
Environmental stability refers to how stable all of the environmental forces are over time. There was a time when certain external forces were considered stable, such as governmental regulations and laws, which could continue for many years, even with new political administrations in place. However, the financial crisis of 2008 and political changes in the United States have added volatility. The high-tech and software industries would be considered unstable because of the relative ease for new software and technology companies to enter and take over an existing market.
Internal Environment
The internal environment is influenced by the jobs and the employees as they relate to those jobs. The employees’ influence on the internal environment is directly related to their level of engagement, or job satisfaction. The job characteristics model, proposed by Richard Hackman and Greg Oldham in 1980, proposes that the right combination of skill variety, task identity, task significance, autonomy, and feedback can lead to high-quality performance, high internal motivation, and high satisfaction.
Skill variety: This is the degree to which the job requires a person to use multiple high-level skills. A department store greeter whose job consists of greeting customers and giving them a shopping cart demonstrates low levels of skill variety, whereas the employee who acts as a cashier, stocks shelves, and manages the inventory of outdoor furniture demonstrates high skill variety.
Task identity: This is the degree to which a person is responsible for completing an identifiable task from start to finish. A graphic designer who creates images for a website might have low task identity. This is because the designer’s work is only part of a larger whole; other designers and coders contribute their work to completing the website. However, a graphic designer who creates a brochure for a client from the idea phase to the final proof will have a high task identity.
Task significance: This is the degree to which a person’s job affects customers or other people’s work. A nurse handling the diverse needs patients in the intensive care unit may score high on significance, whereas new nurses aiding in the same department may feel that they perform only busy work and feel a low level of significance.
Autonomy: This is the degree to which a person can decide how to perform his or her tasks. For instance, a grocery store clerk who is given a list of tasks to complete by the end of the day has greater autonomy than a clerk who is given that same list and told that the list needs to be completed in a particular order, with certain tasks needing to be done by certain times of the day.
Feedback: This is the degree to which people learn whether they are doing their job well. Feedback may come from other people, such as managers, peers, subordinates, and customers, or it may come from the job itself. For instance, a customer service representative might receive feedback from a supervisor, as well as from the customers he or she has tried to help.
Taken all together, the job characteristics model links the task itself to employee motivation. More specifically, a job that is challenging will improve employee motivation but a job that is boring and repetitive will hamper employee motivation. To make a job challenging, managers can ensure the following:
- There are a variety of tasks for the employee to complete.
- The employee feels a sense of autonomy.
- The employee is empowered to make certain decisions.
Growth Cycle
Larry Greiner’s model of growth offers an organic view of the business life cycle. Like people, organizations change as they age and grow. Each growth stage encompasses an evolutionary phase of growth and a revolutionary phase where an organizational crisis will occur, and the business’s ability to handle these crises can determine its future.
Phase 1: Creativity
The creativity phase is marked by early growth of a company due to an emphasis on creating a product or service. The founders of the company are usually technically or entrepreneurially oriented, and they generally disdain management activities. As the complexity of the business increases, the founders will struggle to both grow and manage the business.
Phase 2: Direction
A strong business manager will be brought in to install a functional organizational structure, with formal communication channels and hierarchy. A strong focus on accounting and capital management will drive most business decisions. Top management’s control of all operations diminishes the autonomy that middle-level managers enjoyed, despite their superior knowledge of markets and products.
Phase 3: Delegation
The delegation phase is marked by the application of a decentralized organizational structure. Middle managers are freed up to make decisions and executives monitor the operation and focus on bigger issues, such as mergers or acquisitions. Better coordination of all the operations will be required as the executives feel a loss of control over the middle managers.
Phase 4: Coordination
The coordination of the business structure involves the merging of local units into product groups, a centralization of support functions, and establishment of formal planning procedures. Although resource use becomes more efficient and growth occurs, managers become frustrated with the bureaucratic red tape. The rules and procedures appear more important than productivity and innovation. In turn, corporate staff becomes frustrated with the uncooperative and uninformed managers.
Phase 5: Collaboration
All parts of the organization criticize the resulting bureaucratic structure, and it will take all the key leaders, managers, and employees to collaborate in an attempt to create a better structure. The formal systems and procedures will have to give way to social control and self-discipline. A shift to a problem solving-based approach is needed as teams combine across various business functions and previous systems are simplified.
Contributors and Attributions
- Reading: Factors Impacting Organizational Design. Authored by : David J. Thompson, PhD, and Lumen Learning. License : CC BY: Attribution
- Image: Factors Impacting Organizational Design. Authored by : Lumen Learning. License : CC BY: Attribution


4.2.1E: Isolation and Development
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Social isolation refers to a complete or near-complete lack of contact with society, which can affect all aspects of a person’s life.
Learning Objectives
- Interpret why social isolation can be problematic for a person in society and the importance of social connections
Key Points
- True social isolation is not the same as loneliness. It is often a chronic condition that persists for years and affects all aspects of a person’s existence.
- Emotional isolation is a term used to describe a state of isolation where the individual is emotionally isolated, but may have a well functioning social network.
- Social networks promote good health by providing direct support, encouraging healthy behaviors, and linking people with diffuse social networks that facilitate access to a wide range of resources supportive of health.
- Sociologists debate whether new technologies, such as the Internet and mobile phones, exacerbate social isolation or encourage it.
- A widely-held hypothesis is that social ties link people with diffuse social networks that facilitate access to a wide range of resources supportive of health.
Key Terms
- emotional isolation : Emotional isolation is a term used to describe a state of isolation where the individual is emotionally isolated, but may have a well functioning social network.
- social network : The web of a person’s social, family, and business contacts, who provide material and social resources and opportunities.
- social isolation : Social isolation refers to a complete or near-complete lack of contact with society. It is usually involuntary, making it distinct from isolating tendencies or actions taken by an individual who is seeking to distance himself from society.
Social isolation occurs when members of a social species (like humans) have complete or near-complete lack of contact with society. Social isolation is usually imposed involuntary, not chosen. Social isolation is not the same as loneliness rooted in temporary lack of contact with other humans, nor is it the same as isolating actions that might be consciously undertaken by an individual. A related phenomenon, emotional isolation may occur when individuals are emotionally isolated, even though they may have well-functioning social networks.
While loneliness is often fleeting, true social isolation often lasts for years or decades and tends to be a chronic condition that affects all aspects of a person’s existence and can have serious consequences for health and well being. Socially isolated people have no one to turn to in personal emergencies, no one to confide in during a crisis, and no one against whom to measure their own behavior against or from whom to learn etiquette or socially acceptable behavior. Social isolation can be problematic at any age, although it has different effects for different age groups (that is, social isolation for children may have different effects than social isolation for adults, although both age groups may experience it).
Social isolation can be dangerous because the vitality of individuals’ social relationships affect their health. Social contacts influence individuals’ behavior by encouraging health-promoting behaviors, such as adequate sleep, diet, exercise, and compliance with medical regimens or by discouraging health-damaging behaviors, such as smoking, excessive eating, alcohol consumption, or drug abuse. Socially isolated individuals lack these beneficial influences, as well as lacking a social support network that can provide help and comfort in times of stress and distress. Social relationships can also connect people with diffuse social networks that facilitate access to a wide range of resources supportive of health, such as medical referral networks, access to others dealing with similar problems, or opportunities to acquire needed resources via jobs, shopping, or financial institutions. These effects are different from receiving direct support from a friend; instead, they are based on the ties that close social ties provide to more distant connections.
Sociologists debate whether new technologies, such as the Internet and mobile phones exacerbate social isolation or could help overcome it. With the advent of online social networking communities, people have increasing options for engaging in social activities that do not require real-world physical interaction. Chat rooms, message boards, and other types of communities are now meeting social needs for those who would rather stay home alone, yet still develop communities of online friends.


14.3.1: Indemnity Plan
The oldest type of health insurance plan allowed for the insured person to visit any doctor - including primary care physicians and specialists. The doctor would then bill the insurance company for any visits or procedures done, the insurance company would pay the portion outlined in the plan, and then the doctor would bill the patient for the remainder. This type of insurance plan is also known as a fee-for-service model, and although it used to be the most common type of employer-sponsored healthcare plan, it is almost non-existent today. The problem with this fee-for-service model is that neither the doctors nor the insurance companies were incentivized to control costs and avoid any unnecessary “utilization” or medical procedures. In fact, doctors are incentivized to do just the opposite - perform as many tests and procedures as they could justify, thereby making more money from treating each patient. Similarly, insurance companies could simply pass on the high costs to employers by increasing premiums year over year. Since the 1990s, these plans have become untenable for most employers and beneficiaries (who could be charged large balance-bills), and so very few health insurance organizations offer indemnity plans (Shi & Singh, 2017).


8.1: Non-Right Triangles - Laws of Sines and Cosines
Although right triangles allow us to solve many applications, it is more common to find scenarios where the triangle we are interested in does not have a right angle.
Two radar stations located 20 miles apart both detect a UFO located between them. The angle of elevatio n measured by the first station is 35 degrees. The angle of elevation measured by the second station is 15 degrees. What is the altitude of the UFO?
We see that the triangle formed by the UFO and the two stations is not a right triangle. Of course, in any triangle we could draw an altitude, a perpendicular line from one vertex to the opposite side, forming two right triangles, but it would be nice to have methods for working directly with non-right triangles. In this section, we will expand upon the right triangle trigonometry we learned in Chapter 5, and adapt it to non-right triangles.
Law of Sines
Given an arbitrary non-right triangle, we can drop an altitude, which we temporarily label \(h\), to create two right triangles.
Using the right triangle relationships,
\[\sin (\alpha ) = \dfrac{h}{b}\text{ and }\sin (\beta ) = \dfrac{h}{a}\nonumber\]
Solving both equations for \(h\), we get \(b\sin (\alpha ) = h\) and \(a\sin (\beta ) = h\). Since the \(h\) is the same in both equations, we establish \(b\sin (\alpha ) = a\sin (\beta )\). Dividing both sides by \(ab\), we conclude that
\[\dfrac{\sin (\alpha )}{a} = \dfrac{\sin (\beta )}{b}\nonumber\]
Had we drawn the altitude to be perpendicular to side \(b\) or \(a\), we could similarly establish
\[\dfrac{\sin (\alpha )}{a} = \dfrac{\sin (\gamma )}{c}\text{ and }\dfrac{\sin (\beta )}{b} = \dfrac{\sin (\gamma )}{c}\nonumber\]
Collectively, these relationships are called the Law of Sines.
LAW OF SINES
Given a triangle with angles and sides opposite labeled as shown, the ratio of sine of angle to length of the opposite side will always be equal, or, symbolically,
\[\dfrac{\sin (\alpha )}{a} = \dfrac{\sin (\beta )}{b} = \dfrac{\sin (\gamma )}{c}\]
For clarity, we call side \(a\) the corresponding side of angle \(\alpha\).
Similarly, we call angle \(\alpha\), the corresponding angle of side \(a\).
Likewise for side \(b\) and angle \(\beta\), and for side \(c\) and angle \(\gamma\).
When we use the law of sines, we use any pair of ratios as an equation. In the most straightforward case, we know two angles and one of the corresponding sides.
Example \(\PageIndex{1}\)
In the triangle shown here, solve for the unknown sides and angle.
Solution
Solving for the unknown angle is relatively easy, since the three angles must add to 180 degrees.
From this, we can determine that
\[\gamma = 180^{\circ} - 50^{\circ} - 30^{\circ} = 100^{\circ}\nonumber\]
To find an unknown side, we need to know the corresponding angle, and we also need another known ratio.
Since we know the angle \(50^{\circ}\) and its corresponding side, we can use this for one of the two ratios. To look for side \(b\), we would use its corresponding angle, \(30^{\circ}\).
\[\dfrac{\sin (50^\circ )}{10} = \dfrac{\sin (30^\circ )}{b}\nonumber\]Multiply both sides by \(b\)
\[b\dfrac{\sin (50^\circ )}{10} = \sin (30^\circ )\nonumber\]Divide, or multiply by the reciprocal, to solve for \(b\)
\[b = \sin (30^\circ )\dfrac{10}{\sin (50^\circ )} \approx 6.527\nonumber\]
Similarly, to solve for side \(c\), we set up the equation
\[\dfrac{\sin (50^\circ )}{10} = \dfrac{\sin (100^\circ )}{c}\nonumber\]
\[c = \sin (100^\circ )\dfrac{10}{\sin (50^\circ )} \approx 12.856\nonumber\]
Example \(\PageIndex{2}\)
Find the elevation of the UFO from the beginning of the section.
Solution
To find the elevation of the UFO, we first find the distance from one station to the UFO, such as the side a in the picture, then use right triangle relationships to find the height of the UFO, \(h\).
Since the angles in the triangle add to 180 degrees, the unknown angle of the triangle must be \(180^{\circ} - 15^{\circ} - 35^{\circ} = 130^{\circ}\). This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship:
\[\dfrac{\sin (130^\circ )}{20} = \dfrac{\sin (35^\circ )}{a}\nonumber\]Multiply by \(a\)
\[a\dfrac{\sin (130^\circ )}{20} = \sin (35^\circ )\nonumber\]Divide, or multiply by the reciprocal, to solve for \(a\)
\[a = \dfrac{20\sin (35^\circ )}{\sin (130^\circ )} \approx 14.975\nonumber\]Simplify
The distance from one station to the UFO is about 15 miles. Now that we know a, we can use right triangle relationships to solve for \(h\).
\[\sin (15^\circ ) = \dfrac{opposite}{hypotenuse} = \dfrac{h}{a} = \dfrac{h}14.975\nonumber\]Solve for \(h\)
\[h = 14.975\sin (15^\circ ) \approx 3.876\nonumber\]
The UFO is at an altitude of 3.876 miles.
In addition to solving triangles in which two angles are known, the law of sines can be used to solve for an angle when two sides and one corresponding angle are known.
Example \(\PageIndex{3}\)
In the triangle shown here, solve for the unknown sides and angles.
Solution
In choosing which pair of ratios from the Law of Sines to use, we always want to pick a pair where we know three of the four pieces of information in the equation. In this case, we know the angle \(85^{\circ}\) and its corresponding side, so we will use that ratio. Since our only other known information is the side with length 9, we will use that side and solve for its corresponding angle.
\[\dfrac{\sin (85^\circ )}{12} = \dfrac{\sin (\beta )}{9}\nonumber\]Isolate the unknown
\[\dfrac{9\sin (85^\circ )}{12} = \sin (\beta )\nonumber\]Use the inverse sine to find a first solution
Remember when we use the inverse function that there are two possible answers.
\[\beta = {\sin ^{ - 1}}\left( \dfrac{9\sin (85^\circ )}{12} \right) \approx 48.3438^\circ \nonumber\]By symmetry we find the second possible solution
\[\beta = 180^\circ - 48.3438^\circ = 131.6562^\circ \nonumber\]
In this second case, if \(\beta \approx 132^{\circ}\), then \(\alpha\) would be \(\alpha = 180^{\circ} - 85^{\circ} - 132^{\circ} = -37^{\circ}\), which doesn’t make sense, so the only possibility for this triangle is \(\beta = 48.3438^{\circ}\).
With a second angle, we can now easily find the third angle, since the angles must add to \(180^{\circ}\), so \(\alpha = 180^{\circ} – 85^{\circ} – 48.3438^{\circ} = 46.6562^{\circ}\).
Now that we know \(\alpha\), we can proceed as in earlier examples to find the unknown side \(a\).
\[\dfrac{\sin (85^\circ )}{12} = \dfrac{\sin (46.6562^\circ )}{a}\nonumber\]
\[a = \dfrac{12\sin (46.6562^\circ )}{\sin (85^\circ )} \approx 8.7603\nonumber\]
Notice that in the problem above, when we use Law of Sines to solve for an unknown angle, there can be two possible solutions. This is called the ambiguous case, and can arise when we know two sides and a non-included angle. In the ambiguous case we may find that a particular set of given information can lead to 2, 1 or no solution at all. However, when an accurate picture of the triangle or suitable context is available, we can determine which angle is desired.
Exercise \(\PageIndex{1}\)
Given \(\alpha = 80^\circ ,a = 120,{\rm{ and }}b = 121\), find the corresponding and missing side and angles. If there is more than one possible solution, show both.
- Answer
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\[\frac{\sin (80^{\circ })}{120}=\frac{\sin (\beta )}{121}\nonumber\]
\[1^{\text{st}}\text{ possible solution }\begin{array}{l}\beta = 83.2^\circ \\\gamma = 16.8^\circ \\c = 35.2\end{array}\nonumber\]
\[2^{\text{nd}}\text{ solution }\begin{array}{l}\beta = 96.8^\circ \\\gamma = 3.2^\circ \\c = 6.9\end{array}\nonumber\]
If we were given a picture of the triangle it may be possible to eliminate one of these
Example \(\PageIndex{4}\)
Find all possible triangles if one side has length 4 opposite an angle of \(50^{\circ}\) and a second side has length 10.
Solution
Using the given information, we can look for the angle opposite the side of length 10.
\[\dfrac{\sin (50^\circ )}{4} = \dfrac{\sin (\alpha )}{10}\nonumber\]
\[\sin (\alpha ) = \dfrac{10\sin (50^\circ )}{4} \approx 1.915\nonumber\]
Since the range of the sine function is [-1, 1], it is impossible for the sine value to be 1.915. There are no triangles that can be drawn with the provided dimensions.
Example \(\PageIndex{5}\)
Find all possible triangles if one side has length 6 opposite an angle of \(50^{circ}\) and a second side has length 4.
Solution
Using the given information, we can look for the angle opposite the side of length 4.
\[\dfrac{\sin (50^\circ )}{6} = \dfrac{\sin (\alpha )}{4}\nonumber\]
\[\sin (\alpha ) = \dfrac{4\sin (50^\circ )}{6} \approx 0.511\nonumber\]Use the inverse to find one solution
\[\alpha = {\sin ^{ - 1}}\left( {0.511} \right) \approx 30.710^\circ \nonumber\]By symmetry there is a second possible solution
\[\alpha = 180^\circ - 30.710^\circ = 149.290^\circ \nonumber\]
If we use the angle \(30.710^\circ \), the third angle would be \(180^\circ - 50^\circ - 30.710^\circ = 99.290^\circ \). We can then use Law of Sines again to find the third side.
\[\dfrac{\sin (50^\circ )}{6} = \dfrac{\sin (99.290^\circ )}{c}\nonumber\]Solve for \(c\)
\[c = 7.730\nonumber\]
If we used the angle \(\alpha = 149.290^{\circ}\), the third angle would be \(180^{\circ} – 50^{\circ} – 149.290^{\circ} = –19.29^{\circ}\), which is impossible, so the previous triangle is the only possible one.
Exercise \(\PageIndex{2}\)
Given \(\alpha = 80^\circ ,a = 100,{\rm{ and }}b = 10\) find the missing side and angles. If there is more than one possible solution, show both.
- Answer
-
\[\dfrac{\sin \left( {80^\circ } \right)}{120} = \dfrac{\sin \left( \beta \right)}{10}\quad \beta = 5.65^\circ \text{ or }\beta = 174.35^\circ \nonumber\]; only the first is reasonable.
\[\gamma = 180^\circ - 5.65^\circ - 80^\circ = 94.35^\circ \nonumber\]
\[\frac{\sin (80^{\circ })}{120}=\frac{\sin (94.35 ^{\circ })}{c}\nonumber\]
\[\beta = 5.65^\circ ,\gamma = 94.35^\circ ,c = 101.25\nonumber\]
Law of Cosines
Suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles. How far from port is the boat?
Unfortunately, while the Law of Sines lets us address many non-right triangle cases, it does not allow us to address triangles where the one known angle is included between two known sides, which means it is not a corresponding angle for a known side. For this, we need another tool.
Given an arbitrary non-right triangle, we can drop an altitude, which we te mporarily label \(h\), to create two right triangles. We will divide the base \(b\) into two pieces, one of which we will temporarily label \(x\).
From this picture, we can establish the right triangle relationship
\[\cos (\alpha ) = \dfrac{x}{c}\text{ or equivalently,} x = c\cos \left( \alpha \right)\nonumber\]
Using the Pythagorean Theorem, we can establish
\[{\left( {b - x} \right)^2} + {h^2} = {a^2}\text{ and }{x^2} + {h^2} = {c^2}\nonumber\]
Both of these equations can be solved for \({h^2}\)
\[{h^2} = {a^2} - {\left( {b - x} \right)^2}\text{ and }{h^2} = {c^2} - {x^2}\nonumber\]
Since the left side of each equation is \({h^2}\), the right sides must be equal
\[{c^2} - {x^2} = {a^2} - {\left( {b - x} \right)^2}\nonumber\]Multiply out the right
\[{c^2} - {x^2} = {a^2} - \left( {b^2} - 2bx + {x^2} \right)\nonumber\]Simplify
\[{c^2} - {x^2} = {a^2} - {b^2} + 2bx - {x^2}\nonumber\]
\[{c^2} = {a^2} - {b^2} + 2bx\nonumber\]Isolate \({a^2}\)
\[{a^2} = {c^2} + {b^2} - 2bx\nonumber\]Substitute in \(c\cos (\alpha ) = x\) from above
\[{a^2} = {c^2} + {b^2} - 2bc\cos (\alpha )\nonumber\]
This result is called the Law of Cosines. Depending upon which side we dropped the altitude down from, we could have established this relationship using any of the angles. The important thing to note is that the right side of the equation involves an angle and the sides adjacent to that angle – the left side of the equation involves the side opposite that angle.
LAW OF COSINES
Given a triangle with angles and opposite sides labeled as shown,
\[{a^2} = {c^2} + {b^2} - 2bc\cos (\alpha )\]
\[{b^2} = {a^2} + {c^2} - 2ac\cos (\beta )\]
\[{c^2} = {a^2} + {b^2} - 2ab\cos (\gamma )\]
Notice that if one of the angles of the triangle is 90 degrees, cos(\(90^{circ}\)) = 0, so the formula
\[{c^2} = {a^2} + {b^2} - 2ab\cos (90^\circ )\nonumber\]Simplifies to
\[{c^2} = {a^2} + {b^2}\nonumber\]
You should recognize this as the Pythagorean Theorem. Indeed, the Law of Cosines is sometimes called the Generalized Pythagorean Theorem , since it extends the Pythagorean Theorem to non-right triangles.
Example \(\PageIndex{6}\)
Returning to our question from earlier, suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles. How far from port is the boat?
Solution
The boat turned 20 degrees, so the obtuse angle of the non-right triangle shown in the picture is the supplemental angle, \(180^{\circ} - 20^{\circ} = 160^{\circ}\).
With this, we can utilize the Law of Cosines to find the missing side of the obtuse triangle – the distance from the boat to port.
\[{x^2} = {8^2} + {10^2} - 2(8)(10)\cos (160^\circ )\nonumber\]Evaluate the cosine and simplify
\[{x^2} = 314.3508\nonumber\]Square root both sides
\[x = \sqrt {314.3508} = 17.730\nonumber\]
The boat is 17.73 miles from port.
Example \(\PageIndex{7}\)
Find the unknown side and angles of this triangle.
Solution
Notice that we don’t have both pieces of any side/angle pair, so the Law of Sines would not work with this triangle.
Since we have the angle included between the two known sides, we can turn to Law of Cosines.
Since the left side of any of the Law of Cosines equations involves the side opposite the known angle, the left side in this situation will involve the side x. The other two sides can be used in either order.
\[{x^2} = {10^2} + {12^2} - 2(10)(12)\cos (30^\circ )\nonumber\]Evaluate the cosine
\[{x^2} = {10^2} + {12^2} - 2(10)(12)\dfrac{\sqrt 3 }{2}\nonumber\]Simplify
\[{x^2} = 244 - 120\sqrt 3 \nonumber\]Take the square root
\[x = \sqrt {244 - 120\sqrt 3 } \approx 6.013\nonumber\]
Now that we know an angle and its corresponding side, we can use the Law of Sines to fill in the remaining angles of the triangle. Solving for angle \(\theta\),
\[\dfrac{\sin (30^\circ )}{6.013} = \dfrac{\sin (\theta )}{10}\nonumber\]
\[\sin (\theta ) = \dfrac{10\sin (30^\circ )}{6.013}\nonumber\]Use the inverse sine
\[\theta = {\sin ^{ - 1}}\left( \dfrac{10\sin (30^\circ )}{6.013} \right) \approx 56.256^\circ \nonumber\]
The other possibility for \(\theta\) would be \(\theta = 180^{\circ} – 56.256^{\circ} = 123.744^{\circ}\). In the original picture, \(\theta\) is an acute angle, so 123.744°\(^{\circ}\) doesn’t make sense if we assume the picture is drawn to scale.
Proceeding with \(\theta = 56.256^{\circ}\), we can then find the third angle of the triangle:
\[\phi = 180^\circ - 30^\circ - 56.256^\circ = 93.744^\circ \nonumber\]
In addition to solving for the missing side opposite one known angle, the Law of Cosines allows us to find the angles of a triangle when we know all three sides.
Example \(\PageIndex{8}\)
Solve for the angle \(\alpha\) in the triangle shown.
Solution
Using the Law of Cosines,
\[{20^2} = {18^2} + {25^2} - 2(18)(25)\cos (\alpha )\nonumber\]Simplify
\[400 = 949 - 900\cos (\alpha )\nonumber\]
\[- 549 = - 900\cos (\alpha )\nonumber\]
\[\dfrac{ - 549}{ - 900} = \cos (\alpha )\nonumber\]
\[\alpha = {\cos ^{ - 1}}\left( \dfrac{ - 549}{ - 900} \right) \approx 52.410^\circ \nonumber\]
Exercise \(\PageIndex{3}\)
Given \(\alpha = 25^\circ ,b = 10,{\rm{ and }}c = 20\)find the missing side and angles.
- Answer
-
\[{a^2} = {10^2} + {20^2} - 2(10)(20)\cos (25^\circ )\quad a = 11.725\nonumber\]
\[\dfrac{\sin \left( {25^\circ } \right)}{11.725} = \dfrac{\sin \left( \beta \right)}{10}\quad \beta = 21.1^\circ \text{ or }\beta = 158.9^\circ \nonumber\]
only the first is reasonable since 25° + 158.9° would exceed 180°.
\[\gamma = 180^\circ - 21.1^\circ - 25^\circ = 133.9^\circ \nonumber\]
\[\beta = 21.1^\circ , \gamma = 133.9^\circ , a = 11.725\nonumber\]
Notice that since the inverse cosine can return any angle between 0 and 180 degrees, there will not be any ambiguous cases when using Law of Cosines to find an angle.
Example \(\PageIndex{9}\)
On many cell phones with GPS, an approximate location can be given before the GPS signal is received. This is done by a process called triangulation, which works by using the distance from two known points. Suppose there are two cell phone towers within range of you, located 6000 feet apart along a straight highway that runs east to west, and you know you are north of the highway. Based on the signal delay, it can be determined you are 5050 feet from the first tower, and 2420 feet from the second. Determine your position north and east of the first tower, and determine how far you are from the highway.
Solution
For simplicity, we start by drawing a picture and labeling our given information. Using the Law of Cosines, we can solve f or the angle \(\theta\).
\[{2420^2} = {6000^2} + {5050^2} - 2(5050)(6000)\cos (\theta )\nonumber\]
\[5856400 = 61501500 - 60600000\cos (\theta )\nonumber\]
\[- 554646100 = - 60600000\cos (\theta )\nonumber\]
\[\cos (\theta ) = \dfrac-554646100-60600000 = 0.9183\nonumber\]
\[\theta = {\cos ^{ - 1}}(0.9183) = 23.328^\circ \nonumber\]
Using this angle, we could then use right triangles to find the position of the cell phone relative to the western tower.
\[\cos (23.328^\circ ) = \dfrac{x}5050\nonumber\]
\[x = 5050\cos (23.328^\circ ) \approx 4637.2\text{ feet}\nonumber\]
\[\sin (23.328^\circ ) = \dfrac{y}5050\nonumber\]
\[y = 5050\sin (23.328^\circ ) \approx 1999.8\text{ feet}\nonumber\]
You are 5050 ft from the tower and \(23.328^\circ \) north of east (or, equivalently, 66.672° east of north). Specifically, you are about 4637 feet east and 2000 feet north of the first tower.
Note that if you didn’t know whether you were north or south of the towers, our calculations would have given two possible locations, one north of the highway and one south. To resolve this ambiguity in real world situations, locating a position using triangulation requires a signal from a third tower.
Example \(\PageIndex{10}\)
To measure the height of a hill, a woman measures the angle of elevation to the top of the hill to be 24 degrees. She then moves back 200 feet and measures the angle of elevation to be 22 degrees. Find the height of the hill.
Solution
As with many problems of this nature, it will be helpful to draw a picture.
Notice there are three triangles formed here – the right triangle including the height \(h\) and the 22 degree angle, the right triangle including the height \(h\) and the 24 degree angle, and the (non-right) obtuse triangle including the 200 ft side. Since this is the triangle we have the most information for, we will begin with it. It may seem odd to work with this triangle since it does not include the desired side \(h\), but we don’t have enough information to work with either of the right triangles yet.
We can find the obtuse angle of the triangle, since it and the angle of 24 degrees complete a straight line – a 180 degree angle. The obtuse angle must be \(180^{\circ} - 24^{\circ} = 156^{\circ}\). From this, we can determine that the third angle is \(2^{\circ}\). We know one side is 200 feet, and its corresponding angle is \(2^{\circ}\), so by introducing a temporary variable \(x\) for one of the other sides (as shown below), we can use Law of Sines to solve for this length \(x\).
\[\dfrac{x}{\sin (22^\circ )} = \dfrac{200}{\sin (2^\circ )}\nonumber\]Setting up the Law of Sines
\[x = \sin (22^\circ )\dfrac{200}{\sin (2^\circ )}\nonumber\]isolating the \(x\) value
\[x = 2146.77\text{ ft}\nonumber\]
Now that we know \(x\), we can use right triangle properties to solve for \(h\).
\[\sin (24^\circ ) = \dfrac{\rm{opposite}}{\rm{hypotenuse}} = \dfrac{h}{x} = \dfrac{h}{2146.77}\nonumber\]
\[h = 2146.77\sin (24^\circ ) = 873.17\text{ ft}\nonumber\] The hill is 873 feet high.
Important Topics of This Section
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Law of Sines
Solving for sides
Solving for angles
Ambiguous case, 0, 1 or 2 solutions -
Law of Cosines
Solving for sides
Solving for angles - Generalized Pythagorean Theorem


3.4: Double-Angle, Half-Angle, and Reduction Formulas
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Learning Objectives
- Use double-angle formulas to find exact values
- Use double-angle formulas to verify identities
- Use reduction formulas to simplify an expression
- Use half-angle formulas to find exact values
Bicycle ramps made for competition (see Figure \(\PageIndex{1}\)) must vary in height depending on the skill level of the competitors. For advanced competitors, the angle formed by the ramp and the ground should be \(\theta\) such that \(\tan \theta=\dfrac{5}{3}\). The angle is divided in half for novices. What is the steepness of the ramp for novices? In this section, we will investigate three additional categories of identities that we can use to answer questions such as this one.
Using Double-Angle Formulas to Find Exact Values
In the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The double-angle formulas are a special case of the sum formulas, where \(\alpha=\beta\). Deriving the double-angle formula for sine begins with the sum formula,
\[\sin(\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta\]
If we let \(\alpha=\beta=\theta\), then we have
\[\begin{align*} \sin(\theta+\theta)&= \sin \theta \cos \theta+\cos \theta \sin \theta\\[4pt] \sin(2\theta)&= 2\sin \theta \cos \theta \end{align*}\]
Deriving the double-angle for cosine gives us three options. First, starting from the sum formula, \(\cos(\alpha+\beta)=\cos \alpha \cos \beta−\sin \alpha \sin \beta\),and letting \(\alpha=\beta=\theta\), we have
\[\begin{align*} \cos(\theta+\theta)&= \cos \theta \cos \theta-\sin \theta \sin \theta\\[4pt] \cos(2\theta)&= {\cos}^2 \theta - {\sin}^2 \theta \end{align*}\]
Using the Pythagorean properties, we can expand this double-angle formula for cosine and get two more variations. The first variation is:
\[\begin{align*} \cos(2\theta)&= {\cos}^2 \theta - {\sin}^2 \theta \\[4pt] &= (1-{\sin}^2 \theta)-{\sin}^2 \theta \end{align*}\]
The second variation is:
\[\begin{align*} \cos(2\theta)&= {\cos}^2 \theta-{\sin}^2 \theta\\[4pt] &= {\cos}^2 \theta-(1-{\cos}^2 \theta)\\[4pt] &= 2 {\cos}^2 \theta-1 \end{align*}\]
Similarly, to derive the double-angle formula for tangent, replacing \(\alpha=\beta=\theta\) in the sum formula gives
\[\begin{align*} \tan(\alpha+\beta)&= \dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\\[4pt] \tan(\theta+\theta)&= \dfrac{\tan \theta+\tan \theta}{1-\tan \theta \tan \theta}\\[4pt] \tan(2\theta)&= \dfrac{2\tan \theta}{1-{\tan}^2 \theta} \end{align*}\]
DOUBLE-ANGLE FORMULAS
The double-angle formulas are summarized as follows:
\[\begin{align} \sin(2\theta)&= 2 \sin \theta \cos \theta\\[4pt] \cos(2\theta)&= {\cos}^2 \theta-{\sin}^2 \theta =1-2 {\sin}^2 \theta = 2{\cos}^2 \theta-1\\[4pt] \tan(2\theta)&= \dfrac{2 \tan \theta}{1-{\tan}^2\theta} \end{align}\]
How to: Given the tangent of an angle and the quadrant in which it is located, use the double-angle formulas to find the exact value
- Draw a triangle to reflect the given information.
- Determine the correct double-angle formula.
- Substitute values into the formula based on the triangle.
- Simplify.
Example \(\PageIndex{1}\): Using a Double-Angle Formula to Find the Exact Value Involving Tangent
Given that \(\tan \theta=−\dfrac{3}{4}\) and \(\theta\) is in quadrant II, find the following:
- \(\sin(2\theta)\)
- \(\cos(2\theta)\)
- \(\tan(2\theta)\)
Solution
If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given \(\tan \theta=−\dfrac{3}{4}\),such that \(\theta\) is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because \(\theta\) is in the second quadrant, the adjacent side is on the x -axis and is negative. Use the Pythagorean Theorem to find the length of the hypotenuse:
\[\begin{align*} {(-4)}^2+{(3)}^2&= c^2\\[4pt] 16+9&= c^2\\[4pt] 25&= c^2\\[4pt] c&= 5 \end{align*}\]
Now we can draw a triangle similar to the one shown in Figure \(\PageIndex{2}\).
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Let’s begin by writing the double-angle formula for sine.
\(\sin(2\theta)=2 \sin \theta \cos \theta\)
We see that we to need to find \(\sin \theta\) and \(\cos \theta\). Based on Figure \(\PageIndex{2}\), we see that the hypotenuse equals \(5\), so \(\sin θ=35\), \(\sin θ=35\), and \(\cos θ=−45\). Substitute these values into the equation, and simplify.
Thus,
\[\begin{align*} \sin(2\theta)&= 2\left(\dfrac{3}{5}\right)\left(-\dfrac{4}{5}\right)\\[4pt] &= -\dfrac{24}{25} \end{align*}\]
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Write the double-angle formula for cosine.
\(\cos(2\theta)={\cos}^2 \theta−{\sin}^2 \theta\)
Again, substitute the values of the sine and cosine into the equation, and simplify.
\[\begin{align*} \cos(2\theta)&= {\left(-\dfrac{4}{5}\right)}^2-{\left(\dfrac{3}{5}\right)}^2\\[4pt] &= \dfrac{16}{25}-\dfrac{9}{25}\\[4pt] &= \dfrac{7}{25} \end{align*}\]
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Write the double-angle formula for tangent.
\(\tan(2\theta)=\dfrac{2 \tan \theta}{1−{\tan}^2\theta}\)
In this formula, we need the tangent, which we were given as \(\tan \theta=−\dfrac{3}{4}\). Substitute this value into the equation, and simplify.
\[\begin{align*} \tan(2\theta)&= \dfrac{2\left(-\dfrac{3}{4}\right )}{1-{\left(-\dfrac{3}{4}\right)}^2}\\[4pt]
&= \dfrac{-\dfrac{3}{2}}{1-\dfrac{9}{16}}\\[4pt]
&= -\dfrac{3}{2}\left(\dfrac{16}{7}\right)\\[4pt]
&= -\dfrac{24}{7}
\end{align*}\]
Exercise \(\PageIndex{1}\)
Given \(\sin \alpha=\dfrac{5}{8}\),with \(\theta\) in quadrant I, find \(\cos(2\alpha)\).
- Answer
-
\(\cos(2\alpha)=\dfrac{7}{32}\)
Example \(\PageIndex{2}\): Using the Double-Angle Formula for Cosine without Exact Values
Use the double-angle formula for cosine to write \(\cos(6x)\) in terms of \(cos(3x)\).
Solution
\[\begin{align*} \cos(6x)&= \cos(3x+3x)\\[4pt] &= \cos 3x \cos 3x-\sin 3x \sin 3x\\[4pt] &= {\cos}^2 3x-{\sin}^2 3x \end{align*}\]
Analysis
This example illustrates that we can use the double-angle formula without having exact values. It emphasizes that the pattern is what we need to remember and that identities are true for all values in the domain of the trigonometric function.
Using Double-Angle Formulas to Verify Identities
Establishing identities using the double-angle formulas is performed using the same steps we used to derive the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side.
Example \(\PageIndex{3}\): Using the Double-Angle Formulas to Verify an Identity
Verify the following identity using double-angle formulas:
\[1+\sin(2\theta)={(\sin\theta+\cos\theta)}^2 \nonumber \]
Solution
We will work on the right side of the equal sign and rewrite the expression until it matches the left side.
\[\begin{align*} {(\sin \theta+\cos \theta)}^2&= {\sin}^2 \theta+2 \sin \theta \cos \theta+{\cos}^2 \theta\\[4pt] &= ({\sin}^2 \theta+{\cos}^2 \theta)+2 \sin \theta \cos \theta\\[4pt] &= 1+2 \sin \theta \cos \theta\\[4pt] &= 1+\sin(2\theta) \end{align*}\]
Analysis
This process is not complicated, as long as we recall the perfect square formula from algebra:
\[{(a\pm b)}^2=a^2\pm 2ab+b^2 \nonumber \]
where \(a=\sin \theta\) and \(b=\cos \theta\). Part of being successful in mathematics is the ability to recognize patterns. While the terms or symbols may change, the algebra remains consistent.
Exercise \(\PageIndex{2}\)
Verify the identity: \({\cos}^4 \theta−{\sin}^4 \theta=\cos(2\theta)\).
- Answer
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\({\cos}^4 \theta−{\sin}^4 \theta=({\cos}^2 \theta+{\sin}^2 \theta)({\cos}^2 \theta−{\sin}^2 \theta)=\cos(2\theta)\)
Example \(\PageIndex{4}\): Verifying a Double-Angle Identity for Tangent
Verify the identity: \(\tan(2 \theta)=2\cot \theta−\tan \theta\)
Solution
In this case, we will work with the left side of the equation and simplify or rewrite until it equals the right side of the equation.
\[\begin{align*} \tan(2\theta)&= \dfrac{2 \tan \theta}{1-{\tan}^2 \theta} \qquad \text{Double-angle formula}\\[4pt] &= \dfrac{2 \tan \theta\left (\dfrac{1}{\tan \theta}\right)}{(1-{\tan}^2 \theta)\left (\dfrac{1}{\tan \theta}\right )} \qquad \text{Multiply by a term that results in desired numerator}\\[4pt] &= \dfrac{2}{\dfrac{1}{\tan \theta}-\dfrac{ {\tan}^2 \theta}{\tan \theta}}\\[4pt] &= \dfrac{2}{\cot \theta-\tan \theta} \qquad \text {Use reciprocal identity for } \dfrac{1}{\tan \theta} \end{align*}\]
Analysis
Here is a case where the more complicated side of the initial equation appeared on the right, but we chose to work the left side. However, if we had chosen the left side to rewrite, we would have been working backwards to arrive at the equivalency. For example, suppose that we wanted to show
\[\begin{align*} \dfrac{2\tan \theta}{1-{\tan}^2 \theta}&= \dfrac{2}{\cot \theta-\tan \theta} \\[4pt] \text{Lets work on the right side}\\[4pt] \dfrac{2}{\cot \theta-\tan \theta}&= \frac{2}{\frac{1}{\tan \theta }-\tan \theta }\left ( \frac{\tan \theta }{\tan \theta } \right )\\[4pt] &= \dfrac{2 \tan \theta}{\dfrac{1}{\tan \theta}(\tan \theta)-\tan \theta(\tan \theta)}\\[4pt] &= \dfrac{2 \tan \theta}{1-{\tan}^2 \theta} \end{align*}\]
When using the identities to simplify a trigonometric expression or solve a trigonometric equation, there are usually several paths to a desired result. There is no set rule as to what side should be manipulated. However, we should begin with the guidelines set forth earlier.
Exercise \(\PageIndex{3}\)
Verify the identity: \(\cos(2\theta)\cos \theta={\cos}^3 \theta−\cos \theta {\sin}^2 \theta\).
- Answer
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\(\cos(2 \theta)\cos \theta=({\cos}^2 \theta−{\sin}^2 \theta) \cos \theta={\cos}^3 \theta−\cos \theta {\sin}^2 \theta\)
Use Reduction Formulas to Simplify an Expression
The double-angle formulas can be used to derive the reduction formulas, which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine. They allow us to rewrite the even powers of sine or cosine in terms of the first power of cosine. These formulas are especially important in higher-level math courses, calculus in particular. Also called the power-reducing formulas, three identities are included and are easily derived from the double-angle formulas.
We can use two of the three double-angle formulas for cosine to derive the reduction formulas for sine and cosine. Let’s begin with \(\cos(2\theta)=1−2 {\sin}^2 \theta\). Solve for \({\sin}^2 \theta\):
\[\begin{align*} \cos(2\theta)&= 1-2 {\sin}^2 \theta\\[4pt] 2 {\sin}^2 \theta&= 1-\cos(2\theta)\\[4pt] {\sin}^2 \theta&= \dfrac{1-\cos(2\theta)}{2} \end{align*}\]
Next, we use the formula \(\cos(2\theta)=2 {\cos}^2 \theta−1\). Solve for \({\cos}^2 \theta\):
\[\begin{align*} \cos(2\theta)&= 2 {\cos}^2 \theta-1\\[4pt] 1+\cos(2\theta)&= 2 {\cos}^2 \theta\\[4pt] \dfrac{1+\cos(2\theta)}{2}&= {\cos}^2 \theta \end{align*}\]
The last reduction formula is derived by writing tangent in terms of sine and cosine:
\[\begin{align*} \tan^2 \theta &= \frac{\sin^2 \theta}{\cos^2 \theta} \\[4pt] &= \dfrac{\dfrac{1-\cos(2\theta)}{2}}{\dfrac{1+\cos(2\theta)}{2}} \qquad \tag{Substitute the reduction formulas} \\[4pt] &= \left(\dfrac{1-\cos(2 \theta)}{2}\right)\left(\dfrac{2}{1+\cos(2 \theta)}\right) \\[4pt] &= \dfrac{1-\cos(2 \theta)}{1+\cos(2 \theta)} \end{align*}\]
REDUCTION FORMULAS
The reduction formulas are summarized as follows:
\[{\sin}^2 \theta=\dfrac{1−\cos(2 \theta)}{2}\]
\[{\cos}^2 \theta=\dfrac{1+\cos(2 \theta)}{2}\]
\[{\tan}^2 \theta=\dfrac{1−\cos(2 \theta)}{1+\cos(2 \theta)}\]
Example \(\PageIndex{5}\): Writing an Equivalent Expression Not Containing Powers Greater Than 1
Write an equivalent expression for \({\cos}^4 x\) that does not involve any powers of sine or cosine greater than \(1\).
Solution
We will apply the reduction formula for cosine twice.
\[\begin{align*}
{\cos}^4 x&= {({\cos}^2 x)}^2\\[4pt]
&= {\left(\dfrac{1+\cos(2x)}{2}\right)}^2\qquad \text{Substitute reduction formula}\\[4pt]
&= \dfrac{1}{4}(1+2\cos(2x)+{\cos}^2 (2x))\\[4pt]
&= \dfrac{1}{4}+\dfrac{1}{2} \cos(2x)+\dfrac{1}{4}\left (\dfrac{1+{\cos}^2(2x)}{2}\right )\qquad \text{ Substitute reduction formula for } {\cos}^2 x\\[4pt]
&= \dfrac{1}{4}+\dfrac{1}{2} \cos(2x)+\dfrac{1}{8}+\dfrac{1}{8} \cos(4x)\\[4pt]
&= \dfrac{3}{8}+\dfrac{1}{2} \cos(2x)+\dfrac{1}{8} \cos(4x)
\end{align*}\]
Analysis
The solution is found by using the reduction formula twice, as noted, and the perfect square formula from algebra.
Example \(\PageIndex{6}\): Using the Power-Reducing Formulas to Prove an Identity
Use the power-reducing formulas to prove \({\sin}^3(2x)=\left[ \dfrac{1}{2} \sin(2x) \right] [ 1−\cos(4x) \)
Solution
We will work on simplifying the left side of the equation:
\[\begin{align*} {\sin}^3(2x)&= [\sin(2x)][{\sin}^2(2x)]\\[4pt] &= \sin(2x)\left [\dfrac{1-\cos(4x)}{2}\right ]\qquad \text{Substitute the power-reduction formula.}\\[4pt] &= \sin(2x)\left(\dfrac{1}{2}\right)[1-\cos(4x)]\\[4pt] &= \dfrac{1}{2}[\sin(2x)][1-\cos(4x)] \end{align*}\]
Analysis
Note that in this example, we substituted \(\dfrac{1−\cos(4x)}{2}\) for \({\sin}^2(2x)\). The formula states \({\sin}^2 \theta=\dfrac{1−\cos(2\theta)}{2}\)
We let \(\theta=2x\), so \(2\theta=4x\).
Exercise \(\PageIndex{4}\)
Use the power-reducing formulas to prove that \(10{\cos}^4 x=\dfrac{15}{4}+5\cos(2x)+\dfrac{5}{4}\cos(4x)\).
- Answer
-
\[\begin{align*} 10{\cos}^4 x&= 10{({\cos}^2x)}^2\\[4pt] &= 10{\left[\dfrac{ 1+\cos(2x)}{2} \right]}^2\qquad \text{Substitute reduction formula for } {\cos}^2x\\[4pt] &= \dfrac{10}{4}[1+2\cos(2x)+{\cos}^2(2x)]\\[4pt] &= \dfrac{10}{4}+\dfrac{10}{2}\cos(2x)+\dfrac{10}{4}\left(\dfrac{1+{\cos}^2(2x)}{2}\right)\qquad \text{ Substitute reduction formula for } {\cos}^2 x\\[4pt] &= \dfrac{10}{4}+\dfrac{10}{2} \cos(2x)+\dfrac{10}{8}+\dfrac{10}{8}\cos(4x)\\[4pt] &= \dfrac{30}{8}+5\cos(2x)+\dfrac{10}{8}\cos(4x)\\[4pt] &= \dfrac{15}{4}+5\cos(2x)+\dfrac{5}{4}\cos(4x) \end{align*}\]
Using Half-Angle Formulas to Find Exact Values
The next set of identities is the set of half-angle formulas , which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. If we replace \(\theta\) with \(\dfrac{\alpha}{2}\),the half-angle formula for sine is found by simplifying the equation and solving for \(\sin\left(\dfrac{\alpha}{2}\right)\). Note that the half-angle formulas are preceded by a \(\pm\) sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which \(\dfrac{\alpha}{2}\) terminates.
The half-angle formula for sine is derived as follows:
\[\begin{align*} {\sin}^2 \theta&= \dfrac{1-\cos(2\theta)}{2}\\[4pt] {\sin}^2\left(\dfrac{\alpha}{2}\right)&= \dfrac{1-\left(\cos 2\cdot \dfrac{\alpha}{2}\right)}{2}\\[4pt] &= \dfrac{1-\cos \alpha}{2}\\[4pt] \sin \left(\dfrac{\alpha}{2}\right)&= \pm \sqrt{\dfrac{1-\cos \alpha}{2}} \end{align*}\]
To derive the half-angle formula for cosine, we have
\[\begin{align*} {\cos}^2 \theta&= \dfrac{1+\cos(2\theta)}{2}\\[4pt] {\cos}^2\left(\dfrac{\alpha}{2}\right)&= \dfrac{1+\cos\left(2\cdot \dfrac{\alpha}{2}\right)}{2}\\[4pt] &= \dfrac{1+\cos \alpha}{2}\\[4pt] \cos\left(\dfrac{\alpha}{2}\right)&= \pm \sqrt{\dfrac{1+\cos \alpha}{2}} \end{align*}\]
For the tangent identity, we have
\[\begin{align*} {\tan}^2 \theta&= \dfrac{1-\cos(2\theta)}{1+\cos(2\theta)}\\[4pt] {\tan}^2\left(\dfrac{\alpha}{2}\right)&= \dfrac{1-\cos\left(2\cdot \dfrac{\alpha}{2}\right)}{1+\cos\left(2\cdot \dfrac{\alpha}{2}\right)}\\[4pt] \tan\left(\dfrac{\alpha}{2}\right)&= \pm \sqrt{\dfrac{1-\cos \alpha}{1+\cos \alpha}} \end{align*}\]
HALF-ANGLE FORMULAS
The half-angle formulas are as follows:
\[\begin{align} \sin\left(\dfrac{\alpha}{2}\right)&=\pm \sqrt{\dfrac{1-\cos \alpha}{2}} \label{halfsine} \\[4pt] \cos \left(\dfrac{\alpha}{2} \right) &=\pm \sqrt{\dfrac{1+\cos \alpha}{2}} \\[4pt] \tan\left(\dfrac{\alpha}{2}\right) &=\pm \sqrt{\dfrac{1-\cos \alpha}{1+\cos \alpha}} =\dfrac{\sin \alpha}{1+\cos \alpha} =\dfrac{1-\cos \alpha}{\sin \alpha}\end{align}\]
Example \(\PageIndex{7}\)
Using a Half-Angle Formula to Find the Exact Value of a Sine Function. Find \(\sin(15°)\) using a half-angle formula.
Solution
Since \(15°=\dfrac{30°}{2}\),we use the half-angle formula for sine (Equation \ref{halfsine}):
\[\begin{align*}
\sin \dfrac{30^{\circ}}{2}&= \sqrt{\dfrac{1-\cos 30^{\circ}}{2}}\\[4pt]
&= \sqrt{\dfrac{1-\dfrac{\sqrt{3}}{2}}{2}}\\[4pt]
&= \sqrt{\dfrac{\dfrac{2-\sqrt{3}}{2}}{2}}\\[4pt]
&= \sqrt{\dfrac{2-\sqrt{3}}{4}}\\[4pt]
&= \dfrac{\sqrt{2-\sqrt{3}}}{2}
\end{align*}\]
Remember that we can check the answer with a graphing calculator.
Analysis
Notice that we used only the positive root because \(\sin(15°)\) is positive.
Howto: Given the tangent of an angle and the quadrant in which the angle lies, find the exact values of trigonometric functions of half of the angle.
- Draw a triangle to represent the given information.
- Determine the correct half-angle formula.
- Substitute values into the formula based on the triangle.
- Simplify.
Example \(\PageIndex{8}\): Finding Exact Values Using Half-Angle Identities
Given that \(\tan \alpha=\dfrac{8}{15}\) and \(α\) lies in quadrant III, find the exact value of the following:
- \(\sin\left(\dfrac{\alpha}{2}\right)\)
- \(\cos\left(\dfrac{\alpha}{2}\right)\)
- \(\tan\left(\dfrac{\alpha}{2}\right)\)
Solution
Using the given information, we can draw the triangle shown in Figure \(\PageIndex{3}\). Using the Pythagorean Theorem, we find the hypotenuse to be 17. Therefore, we can calculate \(\sin \alpha=−\dfrac{8}{17}\) and \(\cos \alpha=−\dfrac{15}{17}\).
- Before we start, we must remember that if \(α\) is in quadrant III, then \(180°<\alpha<270°\),so \(\dfrac{180°}{2}<\dfrac{\alpha}{2}<\dfrac{270°}{2}\). This means that the terminal side of \(\dfrac{\alpha}{2}\) is in quadrant II, since \(90°<\dfrac{\alpha}{2}<135°\). To find \(\sin \dfrac{\alpha}{2}\),we begin by writing the half-angle formula for sine. Then we substitute the value of the cosine we found from the triangle in Figure \(\PageIndex{3}\) and simplify. \[\begin{align*} \sin \dfrac{\alpha}{2}&= \pm \sqrt{\dfrac{1-\cos \alpha}{2}}\\[4pt] &= \pm \sqrt{\dfrac{1-(-\dfrac{15}{17})}{2}}\\[4pt] &= \pm \sqrt{\dfrac{\dfrac{32}{17}}{2}}\\[4pt] &= \pm \sqrt{\dfrac{32}{17}\cdot \dfrac{1}{2}}\\[4pt] &= \pm \sqrt{\dfrac{16}{17}}\\[4pt] &= \pm \dfrac{4}{\sqrt{17}}\\[4pt] &= \dfrac{4\sqrt{17}}{17} \end{align*}\] We choose the positive value of \(\sin \dfrac{\alpha}{2}\) because the angle terminates in quadrant II and sine is positive in quadrant II.
- To find \(\cos \dfrac{\alpha}{2}\),we will write the half-angle formula for cosine, substitute the value of the cosine we found from the triangle in Figure \(\PageIndex{3}\), and simplify. \[\begin{align*} \cos \dfrac{\alpha}{2}&= \pm \sqrt{\dfrac{1+\cos \alpha}{2}}\\[4pt] &= \pm \sqrt{\dfrac{1+\left(-\dfrac{15}{17}\right)}{2}}\\[4pt] &= \pm \sqrt{\dfrac{\dfrac{2}{17}}{2}}\\[4pt] &= \pm \sqrt{\dfrac{2}{17}\cdot \dfrac{1}{2}}\\[4pt] &= \pm \sqrt{\dfrac{1}{17}}\\[4pt] &= -\dfrac{\sqrt{17}}{17} \end{align*}\] We choose the negative value of \(\cos \dfrac{\alpha}{2}\) because the angle is in quadrant II because cosine is negative in quadrant II.
- To find \(\tan \dfrac{\alpha}{2}\),we write the half-angle formula for tangent. Again, we substitute the value of the cosine we found from the triangle in Figure \(\PageIndex{3}\) and simplify. \[\begin{align*} \tan \dfrac{\alpha}{2}&= \pm \sqrt{\dfrac{1-\cos \alpha}{1+\cos \alpha}}\\[4pt] &= \pm \sqrt{\dfrac{1-\left(-\dfrac{15}{17}\right)}{1+\left(-\dfrac{15}{17}\right)}}\\[4pt] &= \pm \sqrt{\dfrac{\dfrac{32}{17}}{\dfrac{2}{17}}}\\[4pt] &= \pm \sqrt{\dfrac{32}{2}}\\[4pt] &= -\sqrt{16}\\[4pt] &= -4 \end{align*}\] We choose the negative value of \(\tan \dfrac{\alpha}{2}\) because \(\dfrac{\alpha}{2}\) lies in quadrant II, and tangent is negative in quadrant II.
Exercise \(\PageIndex{5}\)
Given that \(\sin \alpha=−\dfrac{4}{5}\) and \(\alpha\) lies in quadrant IV, find the exact value of \(\cos \left(\dfrac{\alpha}{2}\right)\).
- Answer
-
\(-\dfrac{2}{\sqrt{5}}\)
Example \(\PageIndex{9}\): Finding the Measurement of a Half Angle
Now, we will return to the problem posed at the beginning of the section. A bicycle ramp is constructed for high-level competition with an angle of \(θ\) formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If \(tan θ=53\) for higher-level competition, what is the measurement of the angle for novice competition?
Solution
Since the angle for novice competition measures half the steepness of the angle for the high level competition, and \(\tan \theta=\dfrac{5}{3}\) for high competition, we can find \(\cos \theta\) from the right triangle and the Pythagorean theorem so that we can use the half-angle identities. See Figure \(\PageIndex{4}\).
\[\begin{align*} 3^2+5^2&=34\\[4pt] c&=\sqrt{34} \end{align*}\]
We see that \(\cos \theta=\dfrac{3}{\sqrt{34}}=\dfrac{3\sqrt{34}}{34}\). We can use the half-angle formula for tangent: \(\tan \dfrac{\theta}{2}=\sqrt{\dfrac{1−\cos \theta}{1+\cos \theta}}\). Since \(\tan \theta\) is in the first quadrant, so is \(\tan \dfrac{\theta}{2}\).
\[\begin{align*}
\tan \dfrac{\theta}{2}&= \sqrt{\dfrac{1-\dfrac{3\sqrt{34}}{34}}{1+\dfrac{3\sqrt{34}}{34}}}\\[4pt]
&= \sqrt{\dfrac{\dfrac{34-3\sqrt{34}}{34}}{\dfrac{34+3\sqrt{34}}{34}}}\\[4pt]
&= \sqrt{\dfrac{34-3\sqrt{34}}{34+3\sqrt{34}}}\\[4pt]
&\approx 0.57
\end{align*}\]
We can take the inverse tangent to find the angle: \({\tan}^{−1}(0.57)≈29.7°\). So the angle of the ramp for novice competition is \(≈29.7°\).
Media
Access these online resources for additional instruction and practice with double-angle, half-angle, and reduction formulas.
- Double-Angle Identities
- Half-Angle Identities
Key Equations
| Double-angle formulas |
\(\sin(2\theta)=2\sin \theta \cos \theta\) \(\cos(2\theta)={\cos}^2 \theta−{\sin}^2 \theta\) \(=1−2{\sin}^2 \theta\) \(=2{\cos}^2 \theta−1\) \(\tan(2\theta)=\dfrac{2\tan \theta}{1−{\tan}^2 \theta}\) |
| Reduction formulas |
\({\sin}^2 \theta=\dfrac{1−\cos(2\theta)}{2}\) \({\cos}^2 \theta=\dfrac{1+\cos(2\theta)}{2}\) \({\tan}^2 \theta=\dfrac{1−\cos(2\theta)}{1+\cos(2\theta)}\) |
| Half-angle formulas |
\(\sin \dfrac{\alpha}{2}=\pm \sqrt{\dfrac{1−\cos \alpha}{2}}\) \(\cos \dfrac{\alpha}{2}=\pm \sqrt{\dfrac{1+\cos \alpha}{2}}\) \(\tan \dfrac{\alpha}{2}=\pm \sqrt{\dfrac{1−\cos \alpha}{1+\cos \alpha}}\) \(=\dfrac{\sin \alpha}{1+\cos \alpha}\) \(=\dfrac{1−\cos \alpha}{\sin \alpha}\) |
Key Concepts
- Double-angle identities are derived from the sum formulas of the fundamental trigonometric functions: sine, cosine, and tangent. See Example \(\PageIndex{1}\), Example \(\PageIndex{2}\), Example \(\PageIndex{3}\), and Example \(\PageIndex{4}\).
- Reduction formulas are especially useful in calculus, as they allow us to reduce the power of the trigonometric term. See Example \(\PageIndex{5}\) and Example \(\PageIndex{6}\).
- Half-angle formulas allow us to find the value of trigonometric functions involving half-angles, whether the original angle is known or not. See Example \(\PageIndex{7}\), Example \(\PageIndex{8}\), and Example \(\PageIndex{9}\).


5.17: Short and Long-Run Objectives
- Differentiate between short and long-run objectives that a retailer will set for itself
Earlier, you read that a firm might pursue any number of objectives for any number of reasons. Of course, we cautioned that objective setting is not just stating a goal, ambition or target. It isn’t only about WHAT the firm plans to accomplish, but also HOW the firm can accomplish that goal, thus implying WHY the objective is strategically important.
It is important to note that not all objectives have the same time horizons. More explicitly, firms might expect to achieve some objectives relatively quickly, while others are expected to require more time. Understanding the time horizons for specific objectives is important because a critical element of implementation and control, which is step #5 in the steps of retail strategy planning, is evaluating performance. Particularly important is identifying leading indicators: meaningful factors whose change indicates or predicts future change. That’s an overly complicated way of saying that leading indicators are predictive measurements, which would give a firm confidence that their strategy is on-track.
For example, if a local grocer had a strategic objective to grow revenue by x% by attracting new and re-engaging lapsed shoppers, they might consider a number of intermediate objectives, the attainment of which would be positive leading indicators of the success of the broader strategy. For example, knowing average trip frequency and amount spent per trip, the retailer might set an objective of issuing 5,000 new shopper cards within 6 months. Or, they might set an objective of stimulating re-activation of 10,000 shoppers who had been dormant for >6 months.
Of course, there would be a host of tactics to support these objectives, which would need to be articulated and implemented. But, by tracking new shopper card applications and/or tracking the re-activation of dormant accounts, the data may show the retailer that they’re on-track to grow revenue by x% by attracting new and re-engaging lapsed shoppers. Thus, these intermediate objectives are “performance markers,” or indicators that the organization is meeting its performance goals as it tracks toward long(er)-term strategic goals.
Contributors and Attributions
- Short and Long-Run Objectives. Authored by : Patrick Williams. Provided by : Lumen Learning. License : CC BY: Attribution


3.2.1: Introduction
Like many college students, Chris Johnson ate a lot of ramen noodles, and like many college students, cooking meant microwaving something in his dorm room. At the time, ramen noodle companies did not print microwave instructions on the package, and there was a lack of microwave cookware in which to prepare the popular college food fare.
Many entrepreneurial ideas aim to solve small everyday problems. In his quest to find a better way to prepare ramen noodles, Chris Johnson invented the Rapid Ramen Cooker.
Johnson is now the CEO of Rapid Brands. He founded his company in 2013 with the Rapid Ramen Cooker. Johnson pitched his idea on the TV show Shark Tank . Although he negotiated with investor Mark Cuban for an ownership stake in the company, Johnson wound up not taking the deal. Instead, he made a deal with Walmart to position the Ramen cooker on the retailer’s shelves. Rapid Brands sold more than 4 million Rapid Ramen Cookers in its first two years at Walmart and other retailers. Johnson has since introduced several new products, including cookers for eggs, hot dogs, and brownies.
Finding an entrepreneurial opportunity and persisting until your business becomes successful is not easy and involves risk. But whether you start a new venture because you came up with an idea to solve an everyday problem, or whether you are deliberately searching for an opportunity because you want to start your own business, doing the research beforehand will help mitigate risk and determine the venture’s likelihood of success.


10.8: Real-space correlation coefficient
The real-space correlation coefficient, RSCC, is a measure of the similarity between an electron-density map calculated directly from a structural model and one calculated from experimental data. An advantage of techniques for evaluating goodness of fit in real space is that they can be performed for arbitrary sets of atoms. They are therefore used most often in the refinement of biological macromolecular structures to improve the model fit on a per-residue basis.
\[\text{RSCC} =\dfrac{\sum | \rho_{obs} - \langle \rho_{obs} \rangle| \sum | \rho_{calc} - \langle \rho_{calc} \rangle|}{\sqrt{\sum | \rho_{obs} - \langle \rho_{obs} \rangle|^2 \sum | \rho_{calc} - \langle \rho_{calc} \rangle|^2}} \nonumber \]
( ρ 's are electron density values at grid points that cover the residue in question, obs and calc refer to experimental and model electron density, respectively.)
This metric is similar to the real-space residual RSR, but does not require that the two densities be scaled against each other.


4.9: Add and Subtract Fractions with Different Denominators (Part 2)
Identify and Use Fraction Operations
By now in this chapter, you have practiced multiplying, dividing, adding, and subtracting fractions. The following table summarizes these four fraction operations. Remember: You need a common denominator to add or subtract fractions, but not to multiply or divide fractions.
Fraction multiplication : Multiply the numerators and multiply the denominators.
\[\dfrac{a}{b} \cdot \dfrac{c}{d} = \dfrac{ac}{bd}\]
Fraction division : Multiply the first fraction by the reciprocal of the second.
\[\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \cdot \dfrac{d}{c}\]
Fraction addition : Add the numerators and place the sum over the common denominator. If the fractions have different denominators, first convert them to equivalent forms with the LCD.
\[\dfrac{a}{c} + \dfrac{b}{c} = \dfrac{a + b}{c}\]
Fraction subtraction : Subtract the numerators and place the difference over the common denominator. If the fractions have different denominators, first convert them to equivalent forms with the LCD.
\[\dfrac{a}{c} - \dfrac{a}{c} = \dfrac{a - b}{c}\]
Simplify:
- \(− \dfrac{1}{4} + \dfrac{1}{6}\)
- \(− \dfrac{1}{4} \div \dfrac{1}{6}\)
Solution
First we ask ourselves, “What is the operation?”
- The operation is addition. Do the fractions have a common denominator? No.
| Find the LCD. | |
| Rewrite each fraction as an equivalent fraction with the LCD. | \(- \dfrac{1 \cdot \textcolor{red}{3}}{4 \cdot \textcolor{red}{3}} + \dfrac{1 \cdot \textcolor{red}{2}}{6 \cdot \textcolor{red}{2}} \) |
| Simplify the numerators and denominators. | \(- \dfrac{3}{12} + \dfrac{2}{12} \) |
| Add the numerators and place the sum over the common denominator. | \(- \dfrac{1}{12} \) |
| Check to see if the answer can be simplified. It cannot. |
- The operation is division. We do not need a common denominator.
| To divide fractions, multiply the first fraction by the reciprocal of the second. | \(- \dfrac{1}{4} \cdot \dfrac{6}{1}\) |
| Multiply. | \(- \dfrac{6}{4}\) |
| Simplify. | \(- \dfrac{3}{2} \) |
Simplify:
- \(− \dfrac{3}{4} - \dfrac{1}{6}\)
- \(− \dfrac{3}{4} \cdot \dfrac{1}{6}\)
- Answer a
-
\(-\dfrac{11}{12}\)
- Answer b
-
\(-\dfrac{1}{8}\)
Simplify:
- \(\dfrac{5}{6} \div \left(- \dfrac{1}{4}\right)\)
- \(\dfrac{5}{6} - \left(- \dfrac{1}{4}\right)\)
- Answer a
-
\(-\dfrac{10}{3}\)
- Answer b
-
\(\dfrac{13}{12}\)
Simplify:
- \(\dfrac{5x}{6} - \dfrac{3}{10}\)
- \(\dfrac{5x}{6} \cdot \dfrac{3}{10}\)
Solution
- The operation is subtraction. The fractions do not have a common denominator.
| Rewrite each fraction as an equivalent fraction with the LCD, 30. | \(\dfrac{5x \cdot \textcolor{red}{5}}{6 \cdot \textcolor{red}{5}} - \dfrac{3 \cdot \textcolor{red}{3}}{10 \cdot \textcolor{red}{3}} \) |
| \(\dfrac{25x}{30} - \dfrac{9}{30} \) | |
| Subtract the numerators and place the difference over the common denominator. | \(\dfrac{25x - 9}{30} \) |
- The operation is multiplication; no need for a common denominator.
| To multiply fractions, multiply the numerators and multiply the denominators. | \(\dfrac{5x \cdot 3}{ 6 \cdot 10} \) |
| Rewrite, showing common factors. | \(\dfrac{\cancel{5} \cdot x \cdot \cancel{3}}{2 \cdot \cancel{3} \cdot 2 \cdot \cancel{5}} \) |
| Remove common factors to simplify. | \(\dfrac{x}{4} \) |
Simplify:
- \(\dfrac{3a}{4} - \dfrac{8}{9}\)
- \(\dfrac{3a}{4} \cdot \dfrac{8}{9}\)
- Answer a
-
\(\dfrac{27a-32}{36}\)
- Answer b
-
\(\dfrac{2a}{3}\)
Simplify:
- \(\dfrac{4k}{5} + \dfrac{5}{6}\)
- \(\dfrac{4k}{5} \div \dfrac{5}{6}\)
- Answer a
-
\(\dfrac{24k+25}{30}\)
- Answer b
-
\(\dfrac{24k}{25}\)
Use the Order of Operations to Simplify Complex Fractions
In Multiply and Divide Mixed Numbers and Complex Fractions , we saw that a complex fraction is a fraction in which the numerator or denominator contains a fraction. We simplified complex fractions by rewriting them as division problems. For example,
\[\dfrac{\dfrac{3}{4}}{\dfrac{5}{8}} = \dfrac{3}{4} \div \dfrac{5}{8} \nonumber \]
Now we will look at complex fractions in which the numerator or denominator can be simplified. To follow the order of operations, we simplify the numerator and denominator separately first. Then we divide the numerator by the denominator.
Step 1. Simplify the numerator.
Step 2. Simplify the denominator.
Step 3. Divide the numerator by the denominator.
Step 4. Simplify if possible.
Simplify: \(\dfrac{\left(\dfrac{1}{2}\right)^{2}}{4 + 3^{2}}\).
Solution
| Simplify the numerator. | \(\dfrac{\dfrac{1}{4}}{4 + 3^{2}}\) |
| Simplify the term with the exponent in the denominator. | \(\dfrac{\dfrac{1}{4}}{4 + 9} \) |
| Add the terms in the denominator. | \(\dfrac{\dfrac{1}{4}}{13} \) |
| Divide the numerator by the denominator. | \(\dfrac{1}{4} \div 13 \) |
| Rewrite as multiplication by the reciprocal. | \(\dfrac{1}{4} \cdot \dfrac{1}{13} \) |
| Multiply. | \(\dfrac{1}{52}\) |
Simplify: \(\dfrac{\left(\dfrac{1}{3}\right)^{2}}{2^{3} + 2}\).
- Answer
-
\(\dfrac{1}{90}\)
Simplify: \(\dfrac{1 + 4^{2}}{\left(\dfrac{1}{4}\right)^{2}}\).
- Answer
-
\(272\)
Simplify: \(\dfrac{\dfrac{1}{2} + \dfrac{2}{3}}{\dfrac{3}{4} - \dfrac{1}{6}}\).
Solution
| Rewrite numerator with the LCD of 6 and denominator with LCD of 12. | \(\dfrac{\dfrac{3}{6} + \dfrac{4}{6}}{\dfrac{9}{12} - \dfrac{2}{12}} \) |
| Add in the numerator. Subtract in the denominator. | \(\dfrac{\dfrac{7}{6}}{\dfrac{7}{12}} \) |
| Divide the numerator by the denominator. | \(\dfrac{7}{6} \div \dfrac{7}{12}\) |
| Rewrite as multiplication by the reciprocal. | \(\dfrac{7}{6} \cdot \dfrac{12}{7} \) |
| Rewrite, showing common factors. | \(\dfrac{\cancel{7} \cdot \cancel{6} \cdot 2}{\cancel{6} \cancel{7} \cdot 1} \) |
| Simplify. | \(2 \) |
Simplify: \(\dfrac{\dfrac{1}{3} + \dfrac{1}{2}}{\dfrac{3}{4} - \dfrac{1}{3}}\).
- Answer
-
\(2\)
Simplify: \(\dfrac{\dfrac{2}{3} - \dfrac{1}{2}}{\dfrac{1}{4} + \dfrac{1}{3}}\).
- Answer
-
\(\dfrac{2}{7}\)
Evaluate Variable Expressions with Fractions
We have evaluated expressions before, but now we can also evaluate expressions with fractions. Remember, to evaluate an expression, we substitute the value of the variable into the expression and then simplify.
Evaluate \(x + \dfrac{1}{3}\) when
- \(x = - \dfrac{1}{3}\)
- \(x = - \dfrac{3}{4}\)
Solution
- To evaluate \(x + \dfrac{1}{3}\) when \(x = − \dfrac{1}{3}\), substitute \(− \dfrac{1}{3}\) for \(x\) in the expression.
| Substitute \(\textcolor{red}{- \dfrac{1}{3}}\) for x. | \(\textcolor{red}{- \dfrac{1}{3}} + \dfrac{1}{3} \) |
| Simplify. | \(0 \) |
- To evaluate \(x + \dfrac{1}{3}\) when \(x = − \dfrac{3}{4}\), we substitute \(− \dfrac{3}{4}\) for \(x\) in the expression.
| Substitute \(\textcolor{red}{- \dfrac{3}{4}}\) for x. | \(\textcolor{red}{- \dfrac{1}{3}} + \dfrac{1}{3}\) |
| Rewrite as equivalent fractions with the LCD, 12. | \(- \dfrac{3 \cdot 3}{4 \cdot 3} + \dfrac{1 \cdot 4}{3 \cdot 4} \) |
| Simplify the numerators and denominators. | \(- \dfrac{9}{12} + \dfrac{4}{12} \) |
| Add. | \(- \dfrac{5}{12} \) |
Evaluate \(x + \dfrac{3}{4}\) when:
- \(x = - \dfrac{7}{4}\)
- \(x = - \dfrac{5}{4}\)
- Answer a
-
\(-1\)
- Answer b
-
\(-\dfrac{1}{2}\)
Evaluate \(y + \dfrac{1}{2}\) when:
- \(y = \dfrac{2}{3}\)
- \(y = - \dfrac{3}{4}\)
- Answer a
-
\(\dfrac{7}{6}\)
- Answer b
-
\(-\dfrac{1}{4}\)
Evaluate \(y − \dfrac{5}{6}\) when \(y = - \dfrac{2}{3}\).
Solution
We substitute \(− \dfrac{2}{3}\) for \(y\) in the expression.
| Substitute \(\textcolor{red}{- \dfrac{2}{3}}\) for y. | \(\textcolor{red}{- \dfrac{2}{3}} - \dfrac{5}{6}\) |
| Rewrite as equivalent fractions with the LCD, 6. | \(- \dfrac{4}{6} - \dfrac{5}{6} \) |
| Subtract. | \(- \dfrac{9}{6} \) |
| Simplify. | \(- \dfrac{3}{2} \) |
Evaluate \(y − \dfrac{1}{2}\) when \(y = - \dfrac{1}{4}\).
- Answer
-
\(-\dfrac{3}{4}\)
Evaluate \(x − \dfrac{3}{8}\) when \(x = - \dfrac{5}{2}\).
- Answer
-
\(-\dfrac{23}{8}\)
Evaluate \(2x^2y\) when \(x = \dfrac{1}{4}\) and \(y = − \dfrac{2}{3}\).
Solution
Substitute the values into the expression. In \(2x^2y\), the exponent applies only to \(x\).
| Substitute \(\textcolor{red}{\dfrac{1}{4}}\) for x and \(\textcolor{blue}{- \dfrac{2}{3}}\) for y. | \(2 \left(\textcolor{red}{\dfrac{1}{4}}\right)^{2} \left(\textcolor{blue}{- \dfrac{2}{3}}\right) \) |
| Simplify exponents first. | \(2 \left(\dfrac{1}{16}\right) \left(- \dfrac{2}{3}\right)\) |
| Multiply. The product will be negative. | \(- \dfrac{2}{1} \cdot \dfrac{1}{16} \cdot \dfrac{2}{3} \) |
| Simplify. | \(- \dfrac{4}{48} \) |
| Remove the common factors. | \(- \dfrac{1 \cdot \cancel{4}}{\cancel{4} \cdot 12} \) |
| Simplify. | \(- \dfrac{1}{12} \) |
Evaluate: \(3ab^2\) when \(a = − \dfrac{2}{3}\) and \(b = − \dfrac{1}{2}\).
- Answer
-
\(-\dfrac{1}{2}\)
Evaluate: \(4c^3d\) when \(c = − \dfrac{1}{2}\) and \(d = − \dfrac{4}{3}\).
- Answer
-
\(\dfrac{2}{3}\)
Evaluate: \(\dfrac{p + q}{r}\) when \(p = −4\), \(q = −2\), and \(r = 8\).
Solution
We substitute the values into the expression and simplify.
| Substitute \(\textcolor{red}{-4}\) for p, \(\textcolor{blue}{-2}\) for q and \(\textcolor{magenta}{8}\) for r. | \(\dfrac{\textcolor{red}{-4} + \textcolor{blue}{(-2)}}{\textcolor{magenta}{8}} \) |
| Add in the numerator first. | \(- \dfrac{6}{8}\) |
| Simplify. | \(- \dfrac{3}{4}\) |
Evaluate: \(\dfrac{a + b}{c}\) when \(a = −8\), \(b = −7\), and \(c = 6\).
- Answer
-
\(-\dfrac{5}{2}\)
Evaluate: \(\dfrac{x + y}{z}\) when \(x = 9\), \(y = −18\), and \(z =- 6\).
- Answer
-
\(\dfrac{3}{2}\)
Practice Makes Perfect
Find the Least Common Denominator (LCD)
In the following exercises, find the least common denominator (LCD) for each set of fractions.
- \(\dfrac{2}{3}\) and \(\dfrac{3}{4}\)
- \(\dfrac{3}{4}\) and \(\dfrac{2}{5}\)
- \(\dfrac{7}{12}\) and \(\dfrac{5}{8}\)
- \(\dfrac{9}{16}\) and \(\dfrac{7}{12}\)
- \(\dfrac{13}{30}\) and \(\dfrac{25}{42}\)
- \(\dfrac{23}{30}\) and \(\dfrac{5}{48}\)
- \(\dfrac{21}{35}\) and \(\dfrac{39}{56}\)
- \(\dfrac{18}{35}\) and \(\dfrac{33}{49}\)
- \(\dfrac{2}{3}, \dfrac{1}{6}\) and \(\dfrac{3}{4}\)
- \(\dfrac{2}{3}, \dfrac{1}{4}\) and \(\dfrac{3}{5}\)
Convert Fractions to Equivalent Fractions with the LCD
In the following exercises, convert to equivalent fractions using the LCD.
- \(\dfrac{1}{3}\) and \(\dfrac{1}{4}\), LCD = 12
- \(\dfrac{1}{4}\) and \(\dfrac{1}{5}\), LCD = 20
- \(\dfrac{5}{12}\) and \(\dfrac{7}{8}\), LCD = 24
- \(\dfrac{7}{12}\) and \(\dfrac{5}{8}\), LCD = 24
- \(\dfrac{13}{16}\) and \(- \dfrac{11}{12}\), LCD = 48
- \(\dfrac{11}{16}\) and \(- \dfrac{5}{12}\), LCD = 48
- \(\dfrac{1}{3}, \dfrac{5}{6}\), and \(\dfrac{3}{4}\), LCD = 12
- \(\dfrac{1}{3}, \dfrac{3}{4}\), and \(\dfrac{3}{5}\), LCD = 60
Add and Subtract Fractions with Different Denominators
In the following exercises, add or subtract. Write the result in simplified form.
- \(\dfrac{1}{3} + \dfrac{1}{5}\)
- \(\dfrac{1}{4} + \dfrac{1}{5}\)
- \(\dfrac{1}{2} + \dfrac{1}{7}\)
- \(\dfrac{1}{3} + \dfrac{1}{8}\)
- \(\dfrac{1}{3} - \left(- \dfrac{1}{9}\right)\)
- \(\dfrac{1}{4} - \left(- \dfrac{1}{8}\right)\)
- \(\dfrac{1}{5} - \left(- \dfrac{1}{10}\right)\)
- \(\dfrac{1}{2} - \left(- \dfrac{1}{6}\right)\)
- \(\dfrac{2}{3} + \dfrac{3}{4}\)
- \(\dfrac{3}{4} + \dfrac{2}{5}\)
- \(\dfrac{7}{12} + \dfrac{5}{8}\)
- \(\dfrac{5}{12} + \dfrac{3}{8}\)
- \(\dfrac{7}{12} - \dfrac{9}{16}\)
- \(\dfrac{7}{16} - \dfrac{5}{12}\)
- \(\dfrac{11}{12} - \dfrac{3}{8}\)
- \(\dfrac{5}{8} - \dfrac{7}{12}\)
- \(\dfrac{2}{3} - \dfrac{3}{8}\)
- \(\dfrac{5}{6} - \dfrac{3}{4}\)
- \(− \dfrac{11}{30} + \dfrac{27}{40}\)
- \(− \dfrac{9}{20} + \dfrac{17}{30}\)
- \(− \dfrac{13}{30} + \dfrac{25}{42}\)
- \(− \dfrac{23}{30} + \dfrac{5}{48}\)
- \(− \dfrac{39}{56} - \dfrac{22}{35}\)
- \(− \dfrac{33}{49} - \dfrac{18}{35}\)
- \(- \dfrac{2}{3} - \left(- \dfrac{3}{4}\right)\)
- \(- \dfrac{3}{4} - \left(- \dfrac{4}{5}\right)\)
- \(- \dfrac{9}{16} - \left(- \dfrac{4}{5}\right)\)
- \(- \dfrac{7}{20} - \left(- \dfrac{5}{8}\right)\)
- 1 + \(\dfrac{7}{8}\)
- 1 + \(\dfrac{5}{6}\)
- 1 − \(\dfrac{5}{9}\)
- 1 − \(\dfrac{3}{10}\)
- \(\dfrac{x}{3} + \dfrac{1}{4}\)
- \(\dfrac{y}{2} + \dfrac{2}{3}\)
- \(\dfrac{y}{4} - \dfrac{3}{5}\)
- \(\dfrac{x}{5} - \dfrac{1}{4}\)
Identify and Use Fraction Operations
In the following exercises, perform the indicated operations. Write your answers in simplified form.
- (a) \(\dfrac{3}{4} + \dfrac{1}{6}\) (b) \(\dfrac{3}{4} \div \dfrac{1}{6}\)
- (a) \(\dfrac{2}{3} + \dfrac{1}{6}\) (b) \(\dfrac{2}{3} \div \dfrac{1}{6}\)
- (a) \(- \dfrac{2}{5} - \dfrac{1}{8}\) (b) \(- \dfrac{2}{5} \cdot \dfrac{1}{8}\)
- (a) \(- \dfrac{4}{5} - \dfrac{1}{8}\) (b) \(- \dfrac{4}{5} \cdot \dfrac{1}{8}\)
- (a) \(\dfrac{5n}{6} \div \dfrac{8}{15}\) (b) \(\dfrac{5n}{6} - \dfrac{8}{15}\)
- (a) \(\dfrac{3a}{8} \div \dfrac{7}{12}\) (b) \(\dfrac{3a}{8} - \dfrac{7}{12}\)
- (a) \(\dfrac{9}{10} \cdot \left(− \dfrac{11d}{12}\right)\) (b) \(\dfrac{9}{10} + \left(− \dfrac{11d}{12}\right)\)
- (a) \(\dfrac{4}{15} \cdot \left(− \dfrac{5}{q}\right)\) (b) \(\dfrac{4}{15} + \left(− \dfrac{5}{q}\right)\)
- \(- \dfrac{3}{8} \div \left(- \dfrac{3}{10}\right)\)
- \(- \dfrac{5}{12} \div \left(- \dfrac{5}{9}\right)\)
- \(- \dfrac{3}{8} + \dfrac{5}{12}\)
- \(- \dfrac{1}{8} + \dfrac{7}{12}\)
- \(\dfrac{5}{6} − \dfrac{1}{9}\)
- \(\dfrac{5}{9} − \dfrac{1}{6}\)
- \(\dfrac{3}{8} \cdot \left(− \dfrac{10}{21}\right)\)
- \(\dfrac{7}{12} \cdot \left(− \dfrac{8}{35}\right)\)
- \(− \dfrac{7}{15} - \dfrac{y}{4}\)
- \(− \dfrac{3}{8} - \dfrac{x}{11}\)
- \(\dfrac{11}{12a} \cdot \dfrac{9a}{16}\)
- \(\dfrac{10y}{13} \cdot \dfrac{8}{15y}\)
Use the Order of Operations to Simplify Complex Fractions
In the following exercises, simplify.
- \(\dfrac{\left(\dfrac{1}{5} \right)^{2}}{2 + 3^{2}}\)
- \(\dfrac{\left(\dfrac{1}{3} \right)^{2}}{5 + 2^{2}}\)
- \(\dfrac{2^{3} + 4^{2}}{\left(\dfrac{2}{3}\right)^{2}}\)
- \(\dfrac{3^{3} - 3^{2}}{\left(\dfrac{3}{4}\right)^{2}}\)
- \(\dfrac{\left(\dfrac{3}{5} \right)^{2}}{\left(\dfrac{3}{7}\right)^{2}}\)
- \(\dfrac{\left(\dfrac{3}{4} \right)^{2}}{\left(\dfrac{5}{8}\right)^{2}}\)
- \(\dfrac{2}{\dfrac{1}{3} + \dfrac{1}{5}}\)
- \(\dfrac{5}{\dfrac{1}{4} + \dfrac{1}{3}}\)
- \(\dfrac{\dfrac{2}{3} + \dfrac{1}{2}}{\dfrac{3}{4} - \dfrac{2}{3}}\)
- \(\dfrac{\dfrac{3}{4} + \dfrac{1}{2}}{\dfrac{5}{6} - \dfrac{2}{3}}\)
- \(\dfrac{\dfrac{7}{8} - \dfrac{2}{3}}{\dfrac{1}{2} + \dfrac{3}{8}}\)
- \(\dfrac{\dfrac{3}{4} - \dfrac{3}{5}}{\dfrac{1}{4} + \dfrac{2}{5}}\)
Mixed Practice
In the following exercises, simplify.
- \(\dfrac{1}{2} + \dfrac{2}{3} \cdot \dfrac{5}{12}\)
- \(\dfrac{1}{3} + \dfrac{2}{5} \cdot \dfrac{3}{4}\)
- 1 − \(\dfrac{3}{5} \div \dfrac{1}{10}\)
- 1 − \(\dfrac{5}{6} \div \dfrac{1}{12}\)
- \(\dfrac{2}{3} + \dfrac{1}{6} + \dfrac{3}{4}\)
- \(\dfrac{2}{3} + \dfrac{1}{4} + \dfrac{3}{5}\)
- \(\dfrac{3}{8} - \dfrac{1}{6} + \dfrac{3}{4}\)
- \(\dfrac{2}{5} + \dfrac{5}{8} - \dfrac{3}{4}\)
- 12\(\left(\dfrac{9}{20} − \dfrac{4}{15}\right)\)
- 8\(\left(\dfrac{15}{16} − \dfrac{5}{6}\right)\)
- \(\dfrac{\dfrac{5}{8} + \dfrac{1}{6}}{\dfrac{19}{24}}\)
- \(\dfrac{\dfrac{1}{6} + \dfrac{3}{10}}{\dfrac{14}{30}}\)
- \(\left(\dfrac{5}{9} + \dfrac{1}{6}\right) \div \left(\dfrac{2}{3} − \dfrac{1}{2}\right)\)
- \(\left(\dfrac{3}{4} + \dfrac{1}{6}\right) \div \left(\dfrac{5}{8} − \dfrac{1}{3}\right)\)
In the following exercises, evaluate the given expression. Express your answers in simplified form, using improper fractions if necessary.
-
x + \(\dfrac{1}{2}\) when
- x = \(− \dfrac{1}{8}\)
- x = \(− \dfrac{1}{2}\)
-
x + \(\dfrac{2}{3}\) when
- x = \(− \dfrac{1}{6}\)
- x = \(− \dfrac{5}{3}\)
-
x + \(\left(− \dfrac{5}{6}\right)\) when
- x = \(\dfrac{1}{3}\)
- x = \(− \dfrac{1}{6}\)
-
x + \(\left(− \dfrac{11}{12}\right)\) when
- x = \(\dfrac{11}{12}\)
- x = \(\dfrac{3}{4}\)
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x − \(\dfrac{2}{5}\) when
- x = \(\dfrac{3}{5}\)
- x = \(- \dfrac{3}{5}\)
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x − \(\dfrac{1}{3}\) when
- x = \(\dfrac{2}{3}\)
- x = \(- \dfrac{2}{3}\)
-
\(\dfrac{7}{10}\) − w when
- w = \(\dfrac{1}{2}\)
- w = \(- \dfrac{1}{2}\)
-
\(\dfrac{5}{12}\) − w when
- w = \(\dfrac{1}{4}\)
- w = \(- \dfrac{1}{4}\)
- 4p 2 q when p = \(- \dfrac{1}{2}\) and q = \(\dfrac{5}{9}\)
- 5m 2 n when m = \(- \dfrac{2}{5}\) and n = \(\dfrac{1}{3}\)
- 2x 2 y 3 when x = \(- \dfrac{2}{3}\) and y = \(- \dfrac{1}{2}\)
- 8u 2 v 3 when u = \(- \dfrac{3}{4}\) and v = \(- \dfrac{1}{2}\)
- \(\dfrac{u + v}{w}\) when u = −4, v = −8, w = 2
- \(\dfrac{m + n}{p}\) when m = −6, n = −2, p = 4
- \(\dfrac{a + b}{a - b}\) when a = −3, b = 8
- \(\dfrac{r - s}{r + s}\) when r = 10, s = −5
Everyday Math
- Decorating Laronda is making covers for the throw pillows on her sofa. For each pillow cover, she needs \(\dfrac{3}{16}\) yard of print fabric and \(\dfrac{3}{8}\) yard of solid fabric. What is the total amount of fabric Laronda needs for each pillow cover?
- Baking Vanessa is baking chocolate chip cookies and oatmeal cookies. She needs \(1 \dfrac{1}{4}\) cups of sugar for the chocolate chip cookies, and \(1 \dfrac{1}{8}\) cups for the oatmeal cookies How much sugar does she need altogether?
Writing Exercises
- Explain why it is necessary to have a common denominator to add or subtract fractions.
- Explain how to find the LCD of two fractions.
Self Check
(a) After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
(b) After looking at the checklist, do you think you are well prepared for the next section? Why or why not?
Contributors and Attributions
-
Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (Formerly of Santa Ana College). This content is licensed under Creative Commons Attribution License v4.0 "Download for free at<EMAIL_ADDRESS>."


1.5.7: Communication Research Summary
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Summary
Communication research is important because it focuses on a common goal—to enhance our interactions with others. In this chapter we highlighted how research is done and the basic steps that guide most research projects—identify the topic, write a research question, define key terms, select a methodology, establish a sample, gather and analyze the data, and finally, interpret and share the results. When conducting research, three factors motivate the choices we make: our intended outcomes, theoretical preferences, and methodological preferences. Depending on these factors, research may lead us to greater understanding, allow us to predict or control a communication situation, or create cultural change.
Conceptualizing Communication research can be done more easily by understanding the three broad methodological approaches/paradigms for conducting Communication research: Rhetorical Methodologies, Quantitative Methodologies, and Qualitative Methodologies. The rhetorical approach evaluates messages in various contexts such as political discourse, art, and popular culture. A variety of methods are available such as neo-aristotelian, fantasy theme, narrative, pentadic, feminist, and ideological criticism. Quantitative methods are characterized by counting phenomena and are useful for predicting communication outcomes or comparing cultures and populations. They include experimental research, surveys, content analysis, and meta-analysis. Qualitative methods offer the opportunity to understand human communication as it occurs in a “natural” context rather than a laboratory setting. This is accomplished through ethnography, focus groups, action research, unobtrusive research, historiography, and case studies. While these approaches share similarities, their focus and specific methods are quite different and produce different outcomes. No research methodology or method is better than another. Instead, approaches to Communication research simply reveal different aspects of human communication in action.
Contributions and Affiliations
- Survey of Communication Study. Authored by : Scott T Paynton and Linda K Hahn. Provided by : Humboldt State University. Located at : en.wikibooks.org/wiki/Survey_of_Communication_Study/Preface. License : CC BY-SA: Attribution-ShareAlike


4.5: §29. Comparative and Superlative Forms
Whether they were 1st and 2nd declension words like planus or 3rd declension words like fortis , most Latin adjectives took their comparative and superlative forms according to regular and consistent rules. Thus, “flat, flatter, flattest” was planus , planior , planissimus ; and “strong, stronger, strongest” was fortis , fortior , fortissimus . This feature of Latin adjective morphology has had very little effect on English. However, in the forms planissimus and fortissimus you may recognize the origins of the Italian pianissimo and fortissimo , musical terms (abbreviated as pp and ff ) that mean “very softly” and “very loudly” (“very strongly”).
Several irregular Latin comparative and superlative forms have left their mark on English. In the Indo-European language family, irregular comparison is a feature of the most commonly used adjectives: consider E “good, better, best” or “bad, worse, worst.” A selection of irregular Latin comparatives and superlatives is given here:
| LATIN ADJECTIVES | ENGLISH MEANING | ENGLISH DERIVATIVES |
| bonus, melior, optimus | good, better, best | ameliorate, optimist |
| malus, pejor, pessimus | bad, worse, worst | pejorative, pessimist |
| magnus, major, maximus | big, bigger, biggest | major, majority, maximum |
| parvus, minor, minimus | small, lesser, least | minor, minus, minimum |
| multus, plus, plurimus | much, more, most | plus, plural, plurality |
Read that list through for general interest; you are not expected to commit it to memory. Other irregular Latin comparatives have entered English without change; among these are interior (“inner”), exterior (“outer”), superior (“higher”), inferior (“lower”), prio r (“former”), posterior (“later”), ulterior (“farther”), junior (“younger”), senior (“older”). Their corresponding superlatives include extremus (“outermost”), supremus or summus (“highest”), and ultimus (“farthest”).


14.3: Chain partitioning
In Chapter 6, we discussed Dilworth's Theorem , which told us that for any poset \(\textbf{P}\) of width \(w\), there is a partition of \(\textbf{P}\) into \(w\), but no fewer, chains. However, we were only able to devise an algorithm to find this chain partition (and a maximum antichain) in the special case where \(\textbf{P}\) was an interval order. Now, through the magic of network flows, we will be able to devise an efficient algorithm that works in general for all posets. However, to do so, we will require a slightly more complicated network than we devised in the previous section.
Suppose that the points of our poset \(\textbf{P}\) are \(\{x_1,x_2,…,x_n\}\). We construct a network from \(\textbf{P}\) consisting of the source \(S\), sink \(T\), and two points \(x_i′\) and \(x_i″\) for each point \(x_i\) of \(\textbf{P}\). All edges in our network will have capacity 1. We add edges from \(S\) to \(x_i′\) for \(1≤i≤n\) and from \(x_i″\) to \(T\) for \(1≤i≤n\). Of course, this network wouldn't be too useful, as it has no edges from the single-prime nodes to the double-prime nodes. To resolve this, we add an edge directed from \(x_i′\) to \(x_j″\) if and only if \(x_i<x_j\) in \(\textbf{P}\).
Our running example in this section will be the poset in Figure 14.8.(a) . We'll discuss the points of the poset as \(x_i\) where \(i\) is the number printed next to the point in the diagram.
The first step is to create the network, which we show in Figure 14.8.(b) . In this network, all capacities are 1, edges are directed from bottom to top, the first row of ten vertices is the \(x_i′\) arranged consecutively with \(x_1′\) at the left and \(x_{10}′\) at the right, and the second row of ten vertices is the \(x_i″\) in increasing order of index. To see how this network is constructed, notice that \(x_1<x_3\) in the poset, so we have the directed edge \((x_1′,x_3″)\). Similarly, \(x_4\) is less than \(x_3, x_5\), and \(x_9\) in the poset, leading to three directed edges leaving \(x_4′\) in the network. As a third example, since \(x_9\) is maximal in the poset, there are no directed edges leaving \(x_9′\).
We have not yet seen how we might turn a maximum flow (or minimum cut) in the network we've just constructed into a minimum chain partition or a maximum antichain. It will be easier to see how this works once we have a confirmed maximum flow. Rather than running the labeling algorithm starting from the zero flow, we eyeball a flow, such as the one shown in Figure 14.9 . (Again, we use the convention that thick edges are full, while thin edges are empty.)
When we run the labeling algorithm (using priority \(S,T,x_1′,…,x_{10}′,x_1″,…,x_{10}″)\), we obtain the following list of labels:
\(S:(∗,+,∞)\) \(x_9″:(x_5′,+,1)\) \(x_3′:(S,+,1)\)
\(x_3′:(S,+,1)\) \(x_4″:(x_6′,+,1)\) \(x_1″:(x_7′,+,1)\)
\(x_5′:(S,+,1)\) \(x_5″:(x_6′,+,1)\) \(x_2″:(x_7′,+,1)\)
\(x_6′:(S,+,1)\) \(x_1′:(x_3″,−,1)\) \(x_2′:(x_7′,+,1)\)
\(x_9′:(S,+,1)\) \(x_8′:(x_9″,−,1)\) \(T:(x_2″,+,1)\)
\(x_3″:(x_5′,+,1)\) \(x_7′:(x_4″,−,1)\)
Thus, we find the augmenting path \((S,x_6′,x_4″,x_7′,x_2″,T)\), and the updated flow can be seen in Figure 14.10 .
If we run the labeling algorithm again, the algorithm assigns the labels below, leaving the sink unlabeled.
\(S:(∗,+,∞)\) \(x_5′:(S,+,1)\) \(x_3″:(x_5′,+,1)\) \(x_1′:(x_3″,−,1)\)
\(x_3′:(S,+,1)\) \(x_9′:(S,+,1)\) \(x_9″:(x_5′,+,1)\) \(x_8′:(x_9″,−,1)\)
In Figure 14.10 , the black vertices are those the labeled in the final run, while the gold vertices are the unlabeled vertices.
Now that we've gone over the part you already knew how to do, we need to discuss how to translate this network flow and cut into a chain partition and an antichain. If there is a unit of flow on an edge \((x_i′,x_j″)\), then a good first instinct is to place \(x_i\) and \(x_j\) in the same chain of a chain partition. To be able to do this successfully, of course, we need to ensure that this won't result in two incomparable points being placed in a chain. A way to see that everything works as desired is to think of starting with \((x_i′,x_j″)\) and then looking for flow leaving \(x_j′\). If there is, it goes to a vertex \(x_k″\), so we may add xk to the chain since \(x_i<x_j<x_k\). Continue in this manner until reaching a vertex in the network that does not have any flow leaving it. Then see if \(x_i″\) has flow coming into it. If it does, it's from a vertex \(x_m′\) that can be added since \(x_m<x_i<x_j\).
Let's see how following this process for the flow in Figure 14.10 leads to a chain partition. If we start with \(x_1′\), we see that \((x_1′,x_3″)\) is full, so we place \(x_1\) and \(x_3\) in chain \(C_1\). Since \(x_3′\) has no flow leaving it, there are no greater elements to add to the chain. However, \(x_1″\) has flow in from \(x_2′\), so we add \(x_2\) to \(C_1\). We now see that \(x_2″\) has flow in from \(x_7′\), so now \(C_1=\{x_1,x_2,x_3,x_7\}\). Vertex \(x_7″\) has no flow into it, so the building of the first chain stops. The first vertex we haven't placed into a chain is \(x_4\), so we note that \((x_4′,x_5″)\) is full, placing \(x_4\) and \(x_5\) in chain \(C_2\). We then look from \(x_5′\) and see no flow leaving. However, there is flow into \(x_4″\) from \(x_6′\), so \(x_6\) is added to \(C_2\). There is no flow out of \(x_6″\), so \(C_2=\{x_4,x_5,x_6\}\). Now the first point not in a chain is \(x_8\), so we use the flow from \(x_8′\) to \(x_9″\) to place \(x_8\) and \(x_9\) in chain \(C_3\). Again, no flow out of \(x_9′\), so we look to \(x_8″\), which is receiving flow from \(x_{10}″\). Adding \(x_{10}\) to \(C_3\) gives \(C_3=\{x_8,x_9,x_{10}\}\), and since every point is now in a chain, we may stop.
Even once we see that the above process does in fact generate a chain partition, it is not immediately clear that it's a minimum chain partition. For this, we need to find an antichain of as many points as there are chains in our partition. (In the example we've been using, we need to find a three-element antichain.) This is where tracking the labeled vertices comes in handy. Suppose we have determined a chain \(C=\{x_1<x_2< \cdot \cdot \cdot <x_k\}\) using the network flow. Since \(x_1\) is the minimal element of this chain, there is no flow into \(x_1″\) and hence no flow out of \(x_1″\). Since \(T\) is unlabeled, this must mean that \(x_1″\) is unlabeled. Similarly, \(x_k\) is the maximal element of \(C\), so there is no flow out of \(x_k′\). Thus, \(x_k′\) is labeled. Now considering the sequence of vertices
\(x_k′,x_k″,x_{k−1}′,x_{k−1}″,…,x_2′,x_2″,x_1′,x_1″\),
there must be a place where the vertices switch from being labeled to unlabeled. This must happen with \(x_i′\) labeled and \(x_i″\) unlabeled. To see why, suppose that \(x_i′\) and \(x_i″\) are both unlabeled while \(x_{i+1}′\) and \(x_{i+1}″\) are both labeled. Because \(x_i\) and \(x_{i+1}\) are consecutive in \(C\), there is flow on \((x_i′,x_{i+1}″)\). Therefore, when scanning from \(x_{i+1}″\), the vertex \(x_i′\) would be labeled. For each chain of the chain partition, we then take the first element \(y\) for which \(y′\) is labeled and \(y″\) is unlabeled to form an antichain \(A=\{y_1,…,y_w\}\). To see that \(A\) is an antichain, notice that if \(y_i<y_j\), then \((y_i′,y_j″)\) is an edge in the network. Therefore, the scan from \(y_i′\) would label \(y_j″\). Using this process, we find that a maximum antichain in our example is \(\{x_1,x_5,x_8\}\).


As a nurse providing family-centered care to a cisgender woman, how would you integrate an understanding of different family structures, such as single-parent, dual-career, and extended families, into your nursing interventions?
2 .
Discuss two challenges nurses might face in delivering optimal care for patients and families from diverse cultural backgrounds and religious beliefs and propose strategies to address these challenges.
3 .
Identify two barriers to care for LGBTQIA+ persons and families and discuss nursing strategies to mitigate these barriers and provide culturally competent care.
4 .
How can cultural factors influence family development stages?
5 .
What are three essential components of a family assessment?
6 .
How do cultural influences and individualism/collectivism shape health beliefs?
7 .
What is transcultural nursing, and why is it essential for culturally competent care?
8 .
What is the Four Cs of Culture model, and why is it valuable for nursing practice?
9 .
Who are considered populations at increased risk for poor health outcomes, and how might their experiences with the health-care system be influenced by discrimination, poverty, or other factors?
10 .
What health risk factors affect families at increased risk for poor health outcomes, such as Black Americans or immigrant families, during pregnancy and childbirth?
11 .
What are social determinants of health (SDOH), and how do they contribute to health disparities in the childbearing population?


1.7: Decimal Fractions
Overview
- Decimal Fractions
- Adding and Subtracting Decimal Fractions
- Multiplying Decimal Fractions
- Dividing Decimal Fractions
- Converting Decimal Fractions to Fractions
- Converting Fractions to Decimal Fractions
Decimal Fractions
Fractions are one way we can represent parts of whole numbers. Decimal fractions are another way of representing parts of whole numbers.
A decimal fraction is a fraction in which the denominator is a power of 10.
A decimal fraction uses a decimal point to separate whole parts and fractional parts. Whole parts are written to the left of the decimal point and fractional parts are written to the right of the decimal point. Just as each digit in a whole number has a particular value, so do the digits in decimal positions.
Sample Set A
The following numbers are decimal fractions
57.9
The 9 is in the tenths position. 57.9 = \(10\dfrac{9}{10}\).
6.8014
The 8 is in the tenths position.
The 0 is in the hundredths position.
The 1 is in the thousandths position.
The 4 is in the ten thousandths position.
6.8014 = \(6\dfrac{8014}{10000}\).
Adding and Subtracting Decimal Fractions
Adding/Subtracting Decimal Fractions:
To add or subtract decimal fractions,
- Align the numbers vertically so that the decimal points line up under each other and corresponding decimal positions are in the same column. Add zeros if necessary.
- Add or subtract the numbers as if they were whole numbers.
- Place a decimal point in the resulting sum or difference directly under the other decimal points.
Sample Set B
Find each sum or difference
Multiplying Decimal Fractions
Multiplying Decimal Fractions
To multiply decimals,
- Multiply the numbers as if they were whole numbers.
- Find the sum of the number of decimal places in the factors.
- The number of decimal places in the product is the sum found in step 2.
Sample Set C
Find the following products
Dividing Decimal Fractions
Dividing Decimal Fractions
To divide a decimal by a nonzero decimal,
- Convert the divisor to a whole number by moving the decimal point to the position immediately to the right of the divisor’s last digit.
- Move the decimal point of the dividend to the right the same number of digits it was moved in the divisor.
- Set the decimal point in the quotient by placing a decimal point directly above the decimal point in the dividend.
- Divide as usual.
Sample Set D
Find the following quotients
Converting Decimal Fractions to Fractions
We can convert a decimal fraction to a fraction by reading it and then writing the phrase we have just read. As we read the decimal fraction, we note the place value farthest to the right. We may have to reduce the fraction.
Sample Set E
Convert each decimal fraction to a fraction
Converting Fractions to Decimal Fractions
Sample Set F
Convert the following fractions to decimals. If the division is nonterminating, round to 2 decimal places.
Exercises
For the following problems, perform each indicated operation.
1.84 + 7.11
- Answer
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8.95
15.015 - 6.527
4.904 - 2.67
- Answer
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2.234
156.33 − 24.095
.0012 + 1.53 + 5.1
- Answer
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6.6312
44.98 + 22.8 − 12.76
5.0004 − 3.00004 + 1.6837
- Answer
-
3.68406
1.11 + 12.1212 − 13.131313
4.26 ⋅ 3.2
- Answer
-
13.632
2.97 ⋅ 3.15
23.05 ⋅ 1.1
- Answer
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25.355
5.009 ⋅ 2.106
0.1 ⋅ 3.24
- Answer
-
0.324
100 ⋅ 12.008
1000 ⋅ 12.008
- Answer
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12,008
10,000 ⋅ 12.008
75.642 ÷ 18.01
- Answer
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4.2
51.811 ÷ 1.97
0.0000448 ÷ 0.014
- Answer
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0.0032
0.129516 ÷ 1004
For the following problems, convert each decimal fraction to a fraction.
0.06
- Answer
-
\(\dfrac{3}{50}\)
0.115
3.7
- Answer
-
\(3\dfrac{7}{10}\)
48.1162
712.00004
- Answer
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\(712\dfrac{1}{25000}\)
For the following problems, convert each fraction to a decimal fraction. If the decimal form is nonterminating,round to 3 decimal places.
\(\dfrac{5}{8}\)
\(\dfrac{9}{20}\)
- Answer
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0.45
15 ÷ 22
\(\dfrac{7}{11}\)
- Answer
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0.636
\(\dfrac{2}{9}\)


4.1: Why It Matters- Infancy
Why understand human development during infancy?
Contributors and Attributions
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Welcome to the story of development from infancy through toddlerhood; from birth until about two years of age. Did you ever wonder how babies grow from tiny, helpless infants into well-developed and independent adults? It doesn’t happen overnight, but the process begins right from day one. Infancy is a time when tremendous growth, coordination, and mental development occur. Most infants learn to walk, manipulate objects and can form basic words by the end of infancy. By 5 months a baby will have doubled its birth weight and tripled its birth weight by the first year. By the age of 2, a baby’s weight will have quadrupled!
Researchers have given this part of the life span more attention than any other period, perhaps because changes during this time are so dramatic and so noticeable. We know that much of what happens during these years provides a foundation for one’s life to come, however, it has been a rgued that the significance of development during these years has been overstated (Bruer, 1999). Nevertheless, this is a period of life that contemporary educators, healthcare pro viders, and parents have focused on quite heavily. It is also a time period that can be tricky to study—how do we learn about infant speech when they cannot articulate their thoughts or feelings? For example, through research we know that infants understand speech much earlier than their bodies have matured enough to physically perform it; thus it is evident that their speech patterns develop before the physical growth of their vocal cords is adequate to facilitate speech.
In this module, we will examine the rapid physical growth and development of infants, look at the influences on physical growth and cognitive development, then turn our attention toward emotional and social development in the early years of life.