3.3: Sum and Difference Identities
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Learning Objectives
- Use sum and difference formulas for cosine.
- Use sum and difference formulas for sine.
- Use sum and difference formulas for tangent.
- Use sum and difference formulas for cofunctions.
- Use sum and difference formulas to verify identities.
How can the height of a mountain be measured? What about the distance from Earth to the sun? Like many seemingly impossible problems, we rely on mathematical formulas to find the answers. The trigonometric identities, commonly used in mathematical proofs, have had real-world applications for centuries, including their use in calculating long distances.
The trigonometric identities we will examine in this section can be traced to a Persian astronomer who lived around 950 AD, but the ancient Greeks discovered these same formulas much earlier and stated them in terms of chords. These are special equations or postulates, true for all values input to the equations, and with innumerable applications.
In this section, we will learn techniques that will enable us to solve problems such as the ones presented above. The formulas that follow will simplify many trigonometric expressions and equations. Keep in mind that, throughout this section, the term formula is used synonymously with the word identity .
Using the Sum and Difference Formulas for Cosine
Finding the exact value of the sine, cosine, or tangent of an angle is often easier if we can rewrite the given angle in terms of two angles that have known trigonometric values. We can use the special angles, which we can review in the unit circle shown in Figure \(\PageIndex{2}\).
We will begin with the sum and difference formulas for cosine, so that we can find the cosine of a given angle if we can break it up into the sum or difference of two of the special angles (Table \(\PageIndex{1}\)).
| Sum formula for cosine | \(\cos(\alpha+\beta)=\cos \alpha \cos \beta−\sin \alpha \sin \beta\) |
| Difference formula for cosine | \(\cos(\alpha−\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta\) |
First, we will prove the difference formula for cosines. Let’s consider two points on the unit circle (Figure \(\PageIndex{3}\)). Point \(P\) is at an angle \(\alpha\) from the positive \(x\) - axis with coordinates \((\cos \alpha,\sin \alpha)\) and point \(Q\) is at an angle of \(\beta\) from the positive \(x\) - axis with coordinates \((\cos \beta,\sin \beta)\). Note the measure of angle \(POQ\) is \(\alpha−\beta\).
Label two more points: \(A\) at an angle of \((\alpha−\beta)\) from the positive \(x\) - axis with coordinates \((\cos(\alpha−\beta),\sin(\alpha−\beta))\); and point \(B\) with coordinates \((1,0)\). Triangle \(POQ\) is a rotation of triangle \(AOB\) and thus the distance from \(P\) to \(Q\) is the same as the distance from \(A\) to \(B\).
We can find the distance from \(P\) to \(Q\) using the distance formula.
\[\begin{align*}
d_{PQ}&= \sqrt{{(\cos \alpha - \cos \beta )}^2+{(sin \alpha - sin \beta )}^2}\\[4pt]
&= \sqrt{{\cos}^2 \alpha-2\cos \alpha \cos \beta+{\cos}^2 \beta+{\sin}^2 \alpha-2\sin \alpha \sin \beta+{\sin}^2 \beta} & & \text {Apply Pythagorean identity and simplify.}\\[4pt]
&= \sqrt{({\cos}^2 \alpha+{\sin}^2 \alpha)+({\cos}^2 \beta+{\sin}^2 \beta)-2\cos \alpha \cos \beta-2\sin \alpha \sin \beta}\\[4pt]
&= \sqrt{1+1-2\cos \alpha \cos \beta-2\sin \alpha \sin \beta}\\[4pt]
&= \sqrt{2-2\cos \alpha \cos \beta-2\sin \alpha \sin \beta}\end{align*}\]
Similarly, using the distance formula we can find the distance from \(A\) to \(B\).
\[\begin{align*}
d_{AB}&= \sqrt{{(\cos(\alpha-\beta)-1)}^2+{(\sin(\alpha-\beta)-0)}^2}\\[4pt]
&= \sqrt{{\cos}^2(\alpha-\beta)-2\cos(\alpha-\beta)+1+{\sin}^2(\alpha-\beta)} & & \text {Apply Pythagorean identity and simplify}\\[4pt]
&= \sqrt{({\cos}^2(\alpha-\beta)+{\sin}^2(\alpha-\beta))-2\cos(\alpha-\beta)+1}\\[4pt]
&= \sqrt{1-2\cos(\alpha-\beta)+1}\\[4pt]
&= \sqrt{2-2\cos(\alpha-\beta)} & & \text {Subtract 2 from both sides and divide both sides by −2.}\\[4pt]
\cos \alpha \cos \beta+\sin \alpha \sin \beta&= \cos(\alpha-\beta)
\end{align*}\]
Thus, we have the difference formula for cosine. We can use similar methods to derive the cosine of the sum of two angles.
SUM AND DIFFERENCE FORMULAS FOR COSINE
These formulas can be used to calculate the cosine of sums and differences of angles.
\[\cos(\alpha+\beta)=\cos \alpha \cos \beta−\sin \alpha \sin \beta\]
\[\cos(\alpha−\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta\]
How to: Given two angles, find the cosine of the difference between the angles
- Write the difference formula for cosine.
- Substitute the values of the given angles into the formula.
- Simplify.
Example \(\PageIndex{1}\): Finding the Exact Value Using the Formula for the Cosine of the Difference of Two Angles
Using the formula for the cosine of the difference of two angles, find the exact value of \(\cos\left(\dfrac{5\pi}{4}−\dfrac{\pi}{6}\right)\).
Solution
Begin by writing the formula for the cosine of the difference of two angles. Then substitute the given values.
\[\begin{align*} \cos(\alpha-\beta)&= \cos \alpha \cos \beta+\sin \alpha \sin \beta\\[4pt] \cos\left(\dfrac{5\pi}{4}-\dfrac{\pi}{6}\right)&= \cos\left(\dfrac{5\pi}{4}\right)\cos\left(\dfrac{\pi}{6}\right)+\sin\left(\dfrac{5\pi}{4}\right)\sin\left(\dfrac{\pi}{6}\right)\\[4pt] &= \left(-\dfrac{\sqrt{2}}{2}\right)\left(\dfrac{\sqrt{3}}{2}\right)-\left(\dfrac{\sqrt{2}}{2}\right )\left(\dfrac{1}{2}\right)\\[4pt] &= -\dfrac{\sqrt{6}}{4}-\dfrac{\sqrt{2}}{4}\\[4pt] &= \dfrac{-\sqrt{6}-\sqrt{2}}{4} \end{align*}\]
Keep in mind that we can always check the answer using a graphing calculator in radian mode.
Exercise \(\PageIndex{1}\):
Find the exact value of \(\cos\left(\dfrac{\pi}{3}−\dfrac{\pi}{4}\right)\).
- Answer
-
\(\dfrac{\sqrt{2}+\sqrt{6}}{4}\)
Example \(\PageIndex{2}\): Finding the Exact Value Using the Formula for the Sum of Two Angles for Cosine
Find the exact value of \(\cos(75°)\).
Solution
As \(75°=45°+30°\),we can evaluate \(\cos(75°)\) as \(\cos(45°+30°)\).
\[\begin{align*} \cos(\alpha+\beta)&= \cos \alpha \cos \beta -\sin \alpha \sin \beta\\[4pt] \cos(45^{\circ}+30^{\circ})&= \cos(45^{\circ})\cos(30^{\circ})-\sin(45^{\circ})\sin(30^{\circ})\\[4pt] &= \dfrac{\sqrt{2}}{2}\left(\dfrac{\sqrt{3}}{2}\right)-\dfrac{\sqrt{2}}{2}\left(\dfrac{1}{2}\right)\\[4pt] &= \dfrac{\sqrt{6}}{4}-\dfrac{\sqrt{2}}{4}\\[4pt] &= \dfrac{\sqrt{6}-\sqrt{2}}{4} \end{align*}\]
Keep in mind that we can always check the answer using a graphing calculator in degree mode.
Analysis
Note that we could have also solved this problem using the fact that \( 75°=135°−60°\).
\[\begin{align*} \cos(\alpha-\beta)&= \cos \alpha \cos \beta+\sin \alpha \sin \beta\\[4pt] \cos(135^{\circ}-60^{\circ})&= \cos(135^{\circ})\cos(60^{\circ})+\sin(135^{\circ})\sin(60^{\circ})\\[4pt] &= \left(-\dfrac{\sqrt{2}}{2}\right)\left(\dfrac{1}{2}\right)+\left(\dfrac{\sqrt{2}}{2}\right )\left(\dfrac{\sqrt{3}}{2}\right)\\[4pt] &= -\dfrac{\sqrt{2}}{4}+\dfrac{\sqrt{6}}{4}\\[4pt] &= \dfrac{\sqrt{6}-\sqrt{2}}{4} \end{align*}\]
Exercise \(\PageIndex{2}\)
Find the exact value of \(\cos(105°)\).
- Answer
-
\(\dfrac{\sqrt{2}-\sqrt{6}}{4}\)
Using the Sum and Difference Formulas for Sine
The sum and difference formulas for sine can be derived in the same manner as those for cosine, and they resemble the cosine formulas.
SUM AND DIFFERENCE FORMULAS FOR SINE
These formulas can be used to calculate the sines of sums and differences of angles.
\[\sin(\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta\]
\[\sin(\alpha−\beta)=\sin \alpha \cos \beta−\cos \alpha \sin \beta\]
How to: Given two angles, find the sine of the difference between the angles
- Write the difference formula for sine.
- Substitute the given angles into the formula.
- Simplify.
Example \(\PageIndex{3}\): Using Sum and Difference Identities to Evaluate the Difference of Angles
Use the sum and difference identities to evaluate the difference of the angles and show that part a equals part b.
- \(\sin(45°−30°)\)
- \(\sin(135°−120°)\)
Solution
- Let’s begin by writing the formula and substitute the given angles.
\[\begin{align*} \sin(\alpha-\beta)&= \sin \alpha \cos \beta-\cos \alpha \sin \beta\\[4pt] \sin(45^{\circ}-30^{\circ})&= \sin(45^{\circ})\cos(30^{\circ})-\cos(45^{\circ})\sin(30^{\circ}) \end{align*}\]
Next, we need to find the values of the trigonometric expressions.
\(\sin(45°)=\frac{\sqrt{2}}{2}, \qquad \cos(30°)=\frac{\sqrt{3}}{2}, \qquad \cos(45°)=\frac{\sqrt{2}}{2}, \qquad \sin(30°)=\frac{1}{2}\)
Now we can substitute these values into the equation and simplify.
\[\begin{align*} \sin(45°-30°)&= \dfrac{\sqrt{2}}{2}\left(\dfrac{\sqrt{3}}{2}\right)-\dfrac{\sqrt{2}}{2}\left(\dfrac{1}{2}\right)\\[4pt] &= \dfrac{\sqrt{6}-\sqrt{2}}{4} \end{align*}\]
- Again, we write the formula and substitute the given angles.
\[\begin{align*} \sin(\alpha-\beta)&= \sin \alpha \cos \beta-\cos \alpha \sin \beta\\[4pt] \sin(135^{\circ}-120^{\circ})&= \sin(135^{\circ})\cos(120^{\circ})-\cos(135^{\circ})\sin(120^{\circ}) \end{align*}\]
Next, we find the values of the trigonometric expressions.
\(\sin(135°)=\frac{\sqrt{2}}{2}, \qquad \cos(120°)=-\frac{1}{2}, \qquad \cos(135°)=\frac{\sqrt{2}}{2}, \qquad \sin(120°)=\frac{\sqrt{3}}{2}\)
Now we can substitute these values into the equation and simplify.
\[\begin{align*} \sin(135^{\circ}-120^{\circ})&= \dfrac{\sqrt{2}}{2}\left(-\dfrac{1}{2}\right)-\left(-\dfrac{\sqrt{2}}{2}\right)\left (\dfrac{\sqrt{3}}{2}\right)\\[4pt] &= \dfrac{\sqrt{6}-\sqrt{2}}{4} \end{align*}\]
Example \(\PageIndex{4}\): Finding the Exact Value of an Expression Involving an Inverse Trigonometric Function
Find the exact value of \(\sin\left ({\cos}^{−1}\frac{1}{2}+{\sin}^{−1}\frac{3}{5}\right)\). Then check the answer with a graphing calculator.
Solution
The pattern displayed in this problem is \(\sin(\alpha+\beta)\). Let \(\alpha={\cos}^{−1}\frac{1}{2}\) and \(\beta={\sin}^{−1}\frac{3}{5}\). Then we can write
\[\begin{align*}
\cos \alpha&= \dfrac{1}{2}, \quad 0\leq \alpha\leq \pi\\[4pt]
\sin \beta&= \dfrac{3}{5}, \quad - \dfrac{\pi}{2}\leq \beta\leq \dfrac{\pi}{2}\\[4pt]\end{align*}\]
We will use the Pythagorean identities to find \(\sin \alpha\) and \(\cos \beta\)
\[\begin{align*}
\sin \alpha&= \sqrt{1-{\cos}^2 \alpha}\\[4pt]
&= \sqrt{1-\dfrac{1}{4}}\\[4pt]
&= \sqrt{\dfrac{3}{4}}\\[4pt]
&= \dfrac{\sqrt{3}}{2}\\[4pt]
\cos \beta&= \sqrt{1-{\sin}^2 \beta}\\[4pt]
&= \sqrt{1-\dfrac{9}{25}}\\[4pt]
&= \sqrt{\dfrac{16}{25}}\\[4pt]
&= \dfrac{4}{5}
\end{align*}\]
Using the sum formula for sine,
\[\begin{align*} \sin \left({\cos}^{-1}\tfrac{1}{2}+{\sin}^{-1}\tfrac{3}{5}\right)&= \sin(\alpha+\beta)\\[4pt] &= \sin \alpha \cos \beta+\cos \alpha \sin \beta\\[4pt] &= \dfrac{\sqrt{3}}{2}\cdot \dfrac{4}{5}+\dfrac{1}{2}\cdot \dfrac{3}{5}\\[4pt] &= \dfrac{4\sqrt{3}+3}{10} \end{align*}\]
Using the Sum and Difference Formulas for Tangent
Finding exact values for the tangent of the sum or difference of two angles is a little more complicated, but again, it is a matter of recognizing the pattern.
Finding the sum of two angles formula for tangent involves taking quotient of the sum formulas for sine and cosine and simplifying. Recall, \(\tan x=\dfrac{\sin x}{\cos x}\), when \(\cos x≠0\).
Let’s derive the sum formula for tangent.
\[\begin{align*}
\tan\left (\alpha+\beta\right)&= \dfrac{\sin\left (\alpha+\beta\right )}{\cos(\alpha+\beta)}\\[6pt]
&= \dfrac{\sin \alpha \cos \beta+\cos \alpha \sin \beta}{\cos \alpha \cos \beta-\sin \alpha \sin \beta}\\[6pt]
&= \dfrac{\dfrac{\sin \alpha \cos \beta+\cos \alpha \sin \beta}{\cos \alpha \cos \beta}}{\dfrac{\cos \alpha \cos \beta-\sin \alpha \sin \beta}{\cos \alpha \cos \beta}}\\[6pt]
&= \dfrac{\dfrac{\sin \alpha \cos \beta}{\cos \alpha \cos \beta}+\dfrac{\cos \alpha \sin \beta}{\cos \alpha \cos \beta}}{\dfrac{\cos \alpha \cos \beta}{\cos \alpha \cos \beta}-\dfrac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta}}\\[6pt]
&= \dfrac{\dfrac{\sin \alpha}{\cos \alpha}+\dfrac{\sin \beta}{\cos \beta}}{1-\dfrac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta}}\\[6pt]
&= \dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}
\end{align*}\]
We can derive the difference formula for tangent in a similar way.
SUM AND DIFFERENCE FORMULAS FOR TANGENT
The sum and difference formulas for tangent are:
\[\tan(\alpha+\beta)=\dfrac{\tan \alpha+\tan \beta}{1−\tan \alpha \tan \beta}\]
\[\tan(\alpha-\beta)=\dfrac{\tan \alpha-\tan \beta}{1+\tan \alpha \tan \beta}\]
How to: Given two angles, find the tangent of the sum of the angles
- Write the sum formula for tangent.
- Substitute the given angles into the formula.
- Simplify.
Example \(\PageIndex{5}\): Finding the Exact Value of an Expression Involving Tangent
Find the exact value of \(\tan\left(\dfrac{\pi}{6}+\dfrac{\pi}{4}\right)\).
Solution
Let’s first write the sum formula for tangent and then substitute the given angles into the formula.
\[\begin{align*}
\tan(\alpha+\beta)&= \dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \\[4pt]
\tan\left(\dfrac{\pi}{6}+\dfrac{\pi}{4}\right)&= \dfrac{\tan\left (\dfrac{\pi}{6}\right)+\tan\left (\dfrac{\pi}{4}\right )}{1-\left (\tan\left (\dfrac{\pi}{6}\right ) \right )\left (\tan\left (\dfrac{\pi}{4}\right )\right )}\end{align*}\]
Next, we determine the individual function values within the formula:
\[\tan\left (\dfrac{\pi}{6}\right )= \dfrac{1}{\sqrt{3}}, \quad \text{and} \quad \tan\left (\dfrac{\pi}{4}\right) = 1\]
So we have,
\[\begin{align*}\tan\left (\dfrac{\pi}{6}+\dfrac{\pi}{4}\right)&= \dfrac{\dfrac{1}{\sqrt{3}}+1}{1-\left(\dfrac{1}{\sqrt{3}}\right )(1)}\\[6pt]
&= \dfrac{\dfrac{1+\sqrt{3}}{\sqrt{3}}}{\dfrac{\sqrt{3}-1}{\sqrt{3}}}\\[6pt]
&= \dfrac{1+\sqrt{3}}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}-1}\\[6pt]
&= \dfrac{\sqrt{3}+1}{\sqrt{3}-1}
\end{align*}\]
Exercise \(\PageIndex{5}\):
Find the exact value of \(\tan\left (\dfrac{2\pi}{3}+\dfrac{\pi}{4}\right )\).
- Answer
-
\(\dfrac{1-\sqrt{3}}{1+\sqrt{3}}\)
Example \(\PageIndex{6}\): Finding Multiple Sums and Differences of Angles
Given \(\sin \alpha=\frac{3}{5}, \quad 0<\alpha<\frac{\pi}{2},\) and \(\cos \beta=−\frac{5}{13}, \quad \pi<\beta<\frac{3\pi}{2}\),
find
- \(\sin(\alpha+\beta)\)
- \(\cos(\alpha+\beta)\)
- \(\tan(\alpha+\beta)\)
- \(\tan(\alpha−\beta)\)
Solution
We can use the sum and difference formulas to identify the sum or difference of angles when the ratio of sine, cosine, or tangent is provided for each of the individual angles. To do so, we construct what is called a reference triangle to help find each component of the sum and difference formulas.
- To find \(\sin(\alpha+\beta)\), we begin with \(\sin \alpha=\dfrac{3}{5}\) and \(0<\alpha<\dfrac{\pi}{2}\). The side opposite \(\alpha\) has length 3, the hypotenuse has length 5, and \(\alpha\) is in the first quadrant. See Figure \(\PageIndex{4}\). Using the Pythagorean Theorem, we can find the length of side \(a\):
\[\begin{align*} a^2+3^2&= 5^2\\[4pt] a^2&= 16\\[4pt] a&= 4 \end{align*}\]
Since \(\cos \beta=−\dfrac{5}{13}\) and \(\pi<\beta<\dfrac{3\pi}{2}\),the side adjacent to \(\beta\) is \(−5\),the hypotenuse is \(13\), and \(\beta\) is in the third quadrant. See Figure \(\PageIndex{5}\). Again, using the Pythagorean Theorem, we have
\[\begin{align*} {(-5)}^2+a^2&= {13}^2\\[4pt] 25+a^2&= 169\\[4pt] a^2&= 144\\[4pt] a&= \pm 12 \end{align*}\]
Since \(\beta\) is in the third quadrant, \(a=–12\).
The next step is finding the cosine of \(\alpha\) and the sine of \(\beta\). The cosine ofα α is the adjacent side over the hypotenuse. We can find it from the triangle in Figure \(\PageIndex{5}\): \(\cos \alpha=\dfrac{4}{5}\). We can also find the sine of \(\beta\) from the triangle in Figure \(\PageIndex{5}\), as opposite side over the hypotenuse: \(\sin \beta=−\dfrac{12}{13}\). Now we are ready to evaluate \(\sin(\alpha+\beta)\).
\[\begin{align*} \sin(\alpha+\beta)&= \sin \alpha \cos \beta+\cos \alpha \sin \beta\\[4pt] &= \left(\dfrac{3}{5}\right)\left(-\dfrac{5}{13}\right )+\left (\dfrac{4}{5}\right )\left(-\dfrac{12}{13}\right )\\[4pt] &= -\dfrac{15}{65}-\dfrac{48}{65}\\[4pt] &= -\dfrac{63}{65} \end{align*}\]
- We can find \(\cos(\alpha+\beta)\) in a similar manner. We substitute the values according to the formula.
\[\begin{align*} \cos(\alpha+\beta)&= \cos \alpha \cos \beta-\sin \alpha \sin \beta\\[4pt] &= \left(\dfrac{4}{5}\right)\left(-\dfrac{5}{13}\right)-\left(\dfrac{3}{5}\right )\left(-\dfrac{12}{13}\right)\\[4pt] &= -\dfrac{20}{65}+\dfrac{36}{65}\\[4pt] &= \dfrac{16}{65} \end{align*}\]
- For \(\tan(\alpha+\beta)\),if \(\sin \alpha=\dfrac{3}{5}\) and \(\cos \alpha=\dfrac{4}{5}\), then
\(\tan \alpha=\dfrac{\dfrac{3}{5}}{\dfrac{4}{5}}=\dfrac{3}{4}\)
If \(\sin \beta=−\dfrac{12}{13}\) and \(\cos \beta=−\dfrac{5}{13}\), then
\(\tan \beta=\dfrac{−\dfrac{12}{13}}{−\dfrac{5}{13}}=\dfrac{12}{5}\)
Then,
\[\begin{align*}
\tan(\alpha+\beta)&= \dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\\[6pt]
&= \dfrac{\dfrac{3}{4}+\dfrac{12}{5}}{1-\dfrac{3}{4}\left(\dfrac{12}{5}\right)}\\[6pt]
&= \dfrac{\dfrac{63}{20}}{-\dfrac{16}{20}}\\[6pt]
&= -\dfrac{63}{16}
\end{align*}\]
- To find \(\tan(\alpha−\beta)\), we have the values we need. We can substitute them in and evaluate.
\[\begin{align*}
\tan(\alpha-\beta)&= \dfrac{\tan \alpha-\tan \beta}{1+\tan \alpha \tan \beta}\\[6pt]
&= \dfrac{\dfrac{3}{4}-\dfrac{12}{5}}{1+\dfrac{3}{4}\left(\dfrac{12}{5}\right)}\\[6pt]
&= \dfrac{-\dfrac{33}{20}}{\dfrac{56}{20}}\\[6pt]
&= -\dfrac{33}{56}
\end{align*}\]
Analysis
A common mistake when addressing problems such as this one is that we may be tempted to think that \(\alpha\) and \(\beta\) are angles in the same triangle, which of course, they are not. Also note that
\(\tan(\alpha+\beta)=\sin(\alpha+\beta)\cos(\alpha+\beta)\)
Using Sum and Difference Formulas for Cofunctions
Now that we can find the sine, cosine, and tangent functions for the sums and differences of angles, we can use them to do the same for their cofunctions. You may recall from Right Triangle Trigonometry that, if the sum of two positive angles is \(\frac{\pi}{2}\), those two angles are complements, and the sum of the two acute angles in a right triangle is \(\frac{\pi}{2}\), so they are also complements. In Figure \(\PageIndex{6}\), notice that if one of the acute angles is labeled as \(\theta\), then the other acute angle must be labeled \(\frac{\pi}{2}−\theta\).
Notice also that \(\sin \theta=\cos\left(\frac{\pi}{2}−\theta\right)\),which is opposite over hypotenuse. Thus, when two angles are complimentary, we can say that the sine of \(\theta\) equals the cofunction of the complement of \(\theta\). Similarly, tangent and cotangent are cofunctions, and secant and cosecant are cofunctions.
From these relationships, the cofunction identities are formed. Recall that you first encountered these identities in The Unit Circle: Sine and Cosine Functions .
COFUNCTION IDENTITIES
The cofunction identities are summarized in Table \(\PageIndex{2}\).
| \(\sin \theta=\cos\left(\dfrac{\pi}{2}−\theta\right)\) | \(\cos \theta=\sin\left(\dfrac{\pi}{2}−\theta\right)\) |
| \(\tan \theta=\cot\left(\dfrac{\pi}{2}−\theta\right)\) | \(\cot \theta=\tan\left(\dfrac{\pi}{2}−\theta\right )\) |
| \(\sec \theta=\csc\left (\dfrac{\pi}{2}−\theta\right )\) | \(\csc \theta=\sec\left (\dfrac{\pi}{2}−\theta\right )\) |
Notice that the formulas in the table may also justified algebraically using the sum and difference formulas. For example, using
\(\cos(\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta\)
We can write
\[\begin{align*} \cos\left (\dfrac{\pi}{2}-\theta \right )&= \cos \dfrac{\pi}{2} \cos \theta+\sin \dfrac{\pi}{2} \sin \theta \\[4pt] &=(0)\cos \theta+(1)\sin \theta \\[4pt] &=\sin \theta \end{align*}\]
Example \(\PageIndex{7}\): Finding a Cofunction with the Same Value as the Given Expression
Write \(\tan \dfrac{\pi}{9}\) in terms of its cofunction.
Solution
The cofunction of \(\tan \theta=\cot\left (\dfrac{\pi}{2}−\theta\right )\). Thus,
\[\begin{align*} \tan\left (\dfrac{\pi}{9}\right )&= \cot\left (\dfrac{\pi}{2}-\dfrac{\pi}{9}\right )\\[4pt] &= \cot\left (\dfrac{9\pi}{18}-\dfrac{2\pi}{18}\right )\\[4pt] &= \cot\left (\dfrac{7\pi}{18}\right ) \end{align*}\]
Exercise \(\PageIndex{7}\)
Write \(\sin \dfrac{\pi}{7}\) in terms of its cofunction.
- Answer
-
\(\cos\left (\dfrac{5\pi}{14}\right )\)
Using the Sum and Difference Formulas to Verify Identities
Verifying an identity means demonstrating that the equation holds for all values of the variable. It helps to be very familiar with the identities or to have a list of them accessible while working the problems. Reviewing the general rules presented earlier may help simplify the process of verifying an identity.
How to: Given an identity, verify using sum and difference formulas
- Begin with the expression on the side of the equal sign that appears most complex. Rewrite that expression until it matches the other side of the equal sign. Occasionally, we might have to alter both sides, but working on only one side is the most efficient.
- Look for opportunities to use the sum and difference formulas.
- Rewrite sums or differences of quotients as single quotients.
- If the process becomes cumbersome, rewrite the expression in terms of sines and cosines.
Example \(\PageIndex{8A}\): Verifying an Identity Involving Sine
Verify the identity \(\sin(\alpha+\beta)+\sin(\alpha−\beta)=2\sin \alpha \cos \beta\).
Solution
We see that the left side of the equation includes the sines of the sum and the difference of angles.
\(\sin(\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta\)
\(\sin(\alpha-\beta)=\sin \alpha \cos \beta-\cos \alpha \sin \beta\)
We can rewrite each using the sum and difference formulas.
\[\begin{align*} \sin(\alpha+\beta)+\sin(\alpha-\beta)&= \sin \alpha \cos \beta+\cos \alpha \sin \beta+\sin \alpha \cos \beta-\cos \alpha \sin \beta\\[4pt] &= 2\sin \alpha \cos \beta \end{align*}\]
We see that the identity is verified.
Example \(\PageIndex{8B}\): Verifying an Identity Involving Tangent
Verify the following identity.
\(\dfrac{\sin(\alpha−\beta)}{\cos \alpha \cos \beta}=\tan \alpha−\tan \beta\)
Solution
We can begin by rewriting the numerator on the left side of the equation.
\[\begin{align*} \dfrac{\sin(\alpha-\beta)}{\cos \alpha \cos \beta}&= \dfrac{\sin \alpha \cos \beta-\cos \alpha \sin \beta}{\cos \alpha \cos \beta}\\[4pt] &= \dfrac{\sin \alpha \cos \beta}{\cos \alpha \cos \beta}-\dfrac{\cos \alpha \sin \beta}{\cos \alpha \cos \beta} & & \text{Rewrite using a common denominator}\\[4pt] &= \dfrac{\sin \alpha}{\cos \alpha}-\dfrac{\sin \beta}{\cos \beta} & & \text{Cancel}\\[4pt] &= \tan \alpha-\tan \beta & & \text{Rewrite in terms of tangent} \end{align*}\]
Exercise \(\PageIndex{8}\):
Verify the identity: \(\tan(\pi−\theta)=−\tan \theta\).
- Answer
-
\[\begin{align*} \tan(\pi-\theta)&= \dfrac{\tan(\pi)-\tan \theta}{1+\tan(\pi)\tan \theta}\\[4pt] &= \dfrac{0-\tan \theta}{1+0\cdot \tan \theta}\\[4pt] &= -\tan \theta \end{align*}\]
Example \(\PageIndex{9A}\): Using Sum and Difference Formulas to Solve an Application Problem
Let \(L_1\) and \(L_2\) denote two non-vertical intersecting lines, and let \(θ\) denote the acute angle between \(L_1\) and \(L_2\). See Figure \(\PageIndex{7}\). Show that
\(\tan \theta=\dfrac{m_2-m_1}{1+m_1m_2}\)
where \(m_1\) and \(m_2\) are the slopes of \(L_1\) and \(L_2\) respectively. ( Hint: Use the fact that \(\tan \theta_1=m_1\) and \(\tan \theta_2=m_2\).)
Solution
Using the difference formula for tangent, this problem does not seem as daunting as it might.
\[\begin{align*} \tan \theta&= \tan(\theta_2-\theta_1)\\[4pt] &= \dfrac{\tan \theta_2-\tan \theta_1}{1+\tan \theta_1 \tan \theta_2}\\[4pt] &= \dfrac{m_2-m_1}{1+m_1m_2} \end{align*}\]
Example \(\PageIndex{9B}\): Investigating a Guy-wire Problem
For a climbing wall, a guy-wire \(R\) is attached \(47\) feet high on a vertical pole. Added support is provided by another guy-wire \(S\) attached \(40\) feet above ground on the same pole. If the wires are attached to the ground \(50\) feet from the pole, find the angle \(\alpha\) between the wires. See Figure \(\PageIndex{8}\).
Solution
Let’s first summarize the information we can gather from the diagram. As only the sides adjacent to the right angle are known, we can use the tangent function. Notice that \(\tan \beta=\frac{47}{50}\), and \(\tan(\beta−\alpha)=\frac{40}{50}=\frac{4}{5}\). We can then use difference formula for tangent.
\[\tan(\beta-\alpha) = \dfrac{\tan \beta-\tan \alpha}{1+\tan \beta \tan \alpha}\]
Now, substituting the values we know into the formula, we have,
\[\begin{align*} \dfrac{4}{5}&= \dfrac{\tfrac{47}{50}-\tan \alpha}{1+\tfrac{47}{50}\tan \alpha}\\[4pt]
4\left(1+\tfrac{47}{50}\tan \alpha\right)&= 5\left(\tfrac{47}{50}-\tan \alpha\right)\end{align*}\]
Use the distributive property, and then simplify the functions.
\[\begin{align*} 4(1)+4\left(\tfrac{47}{50}\right)\tan \alpha &= 5\left(\tfrac{47}{50}\right)-5\tan \alpha\\[4pt]
4+3.76\tan \alpha&= 4.7-5\tan \alpha\\[4pt]
5\tan \alpha+3.76\tan \alpha&= 0.7\\[4pt]
8.76 \tan \alpha&= 0.7\\[4pt]
\tan \alpha&\approx 0.07991\\[4pt]
\tan^{-1}(0.07991)&\approx .079741\end{align*}\]
Now we can calculate the angle in degrees.
\[\begin{align*} \alpha &\approx 0.079741\left(\dfrac{180}{\pi}\right)\\[4pt]
&\approx 4.57^{\circ}
\end{align*}\]
Analysis
Occasionally, when an application appears that includes a right triangle, we may think that solving is a matter of applying the Pythagorean Theorem. That may be partially true, but it depends on what the problem is asking and what information is given.
Media
Access these online resources for additional instruction and practice with sum and difference identities.
- Sum and Difference Identities for Cosine
- Sum and Difference Identities for Sine
- Sum and Difference Identities for Tangent
Key Equations
| Sum Formula for Cosine | \(\cos(\alpha+\beta)=\cos \alpha \cos \beta−\sin \alpha \sin \beta\) |
| Difference Formula for Cosine | \(\cos(\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta\) |
| Sum Formula for Sine | \(\sin(\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta\) |
| Difference Formula for Sine | \(\sin(\alpha-\beta)=\sin \alpha \cos \beta-\cos \alpha \sin \beta\) |
| Sum Formula for Tangent | \(\tan(\alpha+\beta)=\dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\) |
| Difference Formula for Tangent | \(\tan (\alpha-\beta)=\dfrac{\tan (\alpha)-\tan (\beta)}{1+\tan (\alpha) \tan (\beta)}\) |
| Cofunction identities |
\(\sin \theta=\cos\left(\dfrac{\pi}{2}-\theta\right)\) \(\cos \theta=\sin\left(\dfrac{\pi}{2}-\theta\right)\) \(\tan \theta=\cot\left(\dfrac{\pi}{2}-\theta\right)\) \(\cot \theta=\tan\left(\dfrac{\pi}{2}-\theta\right)\) \(\sec \theta=\csc\left(\dfrac{\pi}{2}-\theta\right)\) \(\csc \theta=\sec\left(\dfrac{\pi}{2}-\theta\right)\) |
Key Concepts
- The sum formula for cosines states that the cosine of the sum of two angles equals the product of the cosines of the angles minus the product of the sines of the angles. The difference formula for cosines states that the cosine of the difference of two angles equals the product of the cosines of the angles plus the product of the sines of the angles.
- The sum and difference formulas can be used to find the exact values of the sine, cosine, or tangent of an angle. See Example \(\PageIndex{1}\) and Example \(\PageIndex{2}\).
- The sum formula for sines states that the sine of the sum of two angles equals the product of the sine of the first angle and cosine of the second angle plus the product of the cosine of the first angle and the sine of the second angle. The difference formula for sines states that the sine of the difference of two angles equals the product of the sine of the first angle and cosine of the second angle minus the product of the cosine of the first angle and the sine of the second angle. See Example \(\PageIndex{3}\).
- The sum and difference formulas for sine and cosine can also be used for inverse trigonometric functions. See Example \(\PageIndex{4}\).
- The sum formula for tangent states that the tangent of the sum of two angles equals the sum of the tangents of the angles divided by \(1\) minus the product of the tangents of the angles. The difference formula for tangent states that the tangent of the difference of two angles equals the difference of the tangents of the angles divided by \(1\) plus the product of the tangents of the angles. See Example \(\PageIndex{5}\).
- The Pythagorean Theorem along with the sum and difference formulas can be used to find multiple sums and differences of angles. See Example \(\PageIndex{6}\).
- The cofunction identities apply to complementary angles and pairs of reciprocal functions. See Example \(\PageIndex{7}\).
- Sum and difference formulas are useful in verifying identities. See Example \(\PageIndex{8}\) and Example \(\PageIndex{9}\).
- Application problems are often easier to solve by using sum and difference formulas. See Example \(\PageIndex{10}\) and Example \(\PageIndex{11}\).


11.10: Image Formation by Mirrors
Learning Objectives
- Illustrate image formation in a flat mirror.
- Explain with ray diagrams the formation of an image using spherical mirrors.
- Determine focal length and magnification given radius of curvature, distance of object and image.
We only have to look as far as the nearest bathroom to find an example of an image formed by a mirror. Images in flat mirrors are the same size as the object and are located behind the mirror. Like lenses, mirrors can form a variety of images. For example, dental mirrors may produce a magnified image, just as makeup mirrors do. Security mirrors in shops, on the other hand, form images that are smaller than the object. We will use the law of reflection to understand how mirrors form images, and we will find that mirror images are analogous to those formed by lenses.
Figure \(\PageIndex{1}\) helps illustrate how a flat mirror forms an image. Two rays are shown emerging from the same point, striking the mirror, and being reflected into the observer’s eye. The rays can diverge slightly, and both still get into the eye. If the rays are extrapolated backward, they seem to originate from a common point behind the mirror, locating the image. (The paths of the reflected rays into the eye are the same as if they had come directly from that point behind the mirror.) Using the law of reflection—the angle of reflection equals the angle of incidence—we can see that the image and object are the same distance from the mirror. This is a virtual image, since it cannot be projected—the rays only appear to originate from a common point behind the mirror. Obviously, if you walk behind the mirror, you cannot see the image, since the rays do not go there. But in front of the mirror, the rays behave exactly as if they had come from behind the mirror, so that is where the image is situated.
Now let us consider the focal length of a mirror—for example, the concave spherical mirrors in Figure \(\PageIndex{2}\). Rays of light that strike the surface follow the law of reflection. For a mirror that is large compared with its radius of curvature, as in Figure \(\PageIndex{1}\)(a), we see that the reflected rays do not cross at the same point, and the mirror does not have a well-defined focal point. If the mirror had the shape of a parabola, the rays would all cross at a single point, and the mirror would have a well-defined focal point. But parabolic mirrors are much more expensive to make than spherical mirrors. The solution is to use a mirror that is small compared with its radius of curvature, as shown in Figure \(\PageIndex{2}\)(b). (This is the mirror equivalent of the thin lens approximation.) To a very good approximation, this mirror has a well-defined focal point at F that is the focal distance \(f\) from the center of the mirror. The focal length \(f\) of a concave mirror is positive, since it is a converging mirror.
Just as for lenses, the shorter the focal length, the more powerful the mirror; thus, \(P=1 / f\) for a mirror, too. A more strongly curved mirror has a shorter focal length and a greater power. Using the law of reflection and some simple trigonometry, it can be shown that the focal length is half the radius of curvature, or
\[f=\frac{R}{2}, \nonumber \]
where \(R\) is the radius of curvature of a spherical mirror. The smaller the radius of curvature, the smaller the focal length and, thus, the more powerful the mirror.
The convex mirror shown in Figure \(\PageIndex{3}\) also has a focal point. Parallel rays of light reflected from the mirror seem to originate from the point F at the focal distance \(f\) behind the mirror. The focal length and power of a convex mirror are negative, since it is a diverging mirror.
Ray tracing is as useful for mirrors as for lenses. The rules for ray tracing for mirrors are based on the illustrations just discussed:
- Principal Ray 1 : a ray approaching a concave converging mirror parallel to its axis is reflected through the focal point F of the mirror on the same side. (See rays 1 and 3 in Figure \(\PageIndex{2}\)(b).); a ray approaching a convex diverging mirror parallel to its axis is reflected so that it seems to come from the focal point F behind the mirror. (See rays 1 and 3 in Figure \(\PageIndex{3}\).)
- Principal Ray 2 : any ray striking the center of a mirror is followed by applying the law of reflection; it makes the same angle with the axis when leaving as when approaching. (See ray 2 in Figure \(\PageIndex{4}\).)
- Principal Ray 3 : a ray approaching a concave converging mirror through its focal point is reflected parallel to its axis. (The reverse of rays 1 and 3 in Figure \(\PageIndex{2}\).); a ray approaching a convex diverging mirror by heading toward its focal point on the opposite side is reflected parallel to the axis. (The reverse of rays 1 and 3 in Figure \(\PageIndex{3}\).) The third principal ray is optional and may be used to verify the accuracy of image location.
We will use ray tracing to illustrate how images are formed by mirrors, and we can use ray tracing quantitatively to obtain numerical information. But since we assume each mirror is small compared with its radius of curvature, we can use the thin lens equations for mirrors just as we did for lenses.
Consider the situation shown in Figure \(\PageIndex{4}\), concave spherical mirror reflection, in which an object is placed farther from a concave (converging) mirror than its focal length. That is, \(f\) is positive and \(d_{\mathrm{o}}>f\), so that we may expect an image similar to the case 1 real image formed by a converging lens. Ray tracing in Figure \(\PageIndex{4}\) shows that the rays from a common point on the object all cross at a point on the same side of the mirror as the object. Thus a real image can be projected onto a screen placed at this location. The image distance is positive, and the image is inverted, so its magnification is negative. This is a case 1 image for mirrors . It differs from the case 1 image for lenses only in that the image is on the same side of the mirror as the object. It is otherwise identical.
Example \(\PageIndex{1}\): A Concave Reflector
Electric room heaters use a concave mirror to reflect infrared (IR) radiation from hot coils. Note that IR follows the same law of reflection as visible light. Given that the mirror has a radius of curvature of 50.0 cm and produces an image of the coils 3.00 m away from the mirror, where are the coils?
Strategy and Concept
We are given that the concave mirror projects a real image of the coils at an image distance \(d_{\mathrm{i}}=3.00 \mathrm{~m}\). The coils are the object, and we are asked to find their location—that is, to find the object distance \(d_{\mathrm{o}}\). We are also given the radius of curvature of the mirror, so that its focal length is \(f=R / 2=25.0 \mathrm{~cm}\) (positive since the mirror is concave or converging). Assuming the mirror is small compared with its radius of curvature, we can use the thin lens equations, to solve this problem.
Solution
Since \(d_{\mathrm{i}}\) and \(f\) are known, thin lens equation can be used to find \(d_{\mathrm{o}}\):
\[\frac{1}{d_{\mathrm{o}}}+\frac{1}{d_{\mathrm{i}}}=\frac{1}{f}. \nonumber\]
Rearranging to isolate \(d_{\mathrm{o}}\) gives
\[\frac{1}{d_{\mathrm{o}}}=\frac{1}{f}-\frac{1}{d_{\mathrm{i}}}. \nonumber\]
Entering known quantities gives a value for \(1 / d_{\mathrm{o}}\):
\[\frac{1}{d_{\mathrm{o}}}=\frac{1}{0.250 \mathrm{~m}}-\frac{1}{3.00 \mathrm{~m}}=\frac{3.667}{\mathrm{~m}}. \nonumber\]
This must be inverted to find \(d_{\mathrm{o}}\):
\[d_{\mathrm{o}}=\frac{1 \mathrm{~m}}{3.667}=27.3 \mathrm{~cm}. \nonumber\]
Discussion
Note that the object (the filament) is farther from the mirror than the mirror’s focal length. This is a case 1 image (\(d_{\mathrm{o}}>f\) and \(f\) positive), consistent with the fact that a real image is formed. You will get the most concentrated thermal energy directly in front of the mirror and 3.00 m away from it. Generally, this is not desirable, since it could cause burns. Usually, you want the rays to emerge parallel, and this is accomplished by having the filament at the focal point of the mirror.
Note that the filament here is not much farther from the mirror than its focal length and that the image produced is considerably farther away. This is exactly analogous to a slide projector. Placing a slide only slightly farther away from the projector lens than its focal length produces an image significantly farther away. As the object gets closer to the focal distance, the image gets farther away. In fact, as the object distance approaches the focal length, the image distance approaches infinity and the rays are sent out parallel to one another.
Example \(\PageIndex{2}\): Solar Electric Generating System
One of the solar technologies used today for generating electricity is a device (called a parabolic trough or concentrating collector) that concentrates the sunlight onto a blackened pipe that contains a fluid. This heated fluid is pumped to a heat exchanger, where its heat energy is transferred to another system that is used to generate steam—and so generate electricity through a conventional steam cycle. Figure \(\PageIndex{5}\) shows such a working system in southern California. Concave mirrors are used to concentrate the sunlight onto the pipe. The mirror has the approximate shape of a section of a cylinder. For the problem, assume that the mirror is exactly one-quarter of a full cylinder.
- If we wish to place the fluid-carrying pipe 40.0 cm from the concave mirror at the mirror’s focal point, what will be the radius of curvature of the mirror?
- Per meter of pipe, what will be the amount of sunlight concentrated onto the pipe, assuming the insolation (incident solar radiation) is \(0.900 \mathrm{~kW} / \mathrm{m}^{2}\)?
- If the fluid-carrying pipe has a 2.00-cm diameter, what will be the temperature increase of the fluid per meter of pipe over a period of one minute? Assume all the solar radiation incident on the reflector is absorbed by the pipe, and that the fluid is mineral oil.
Strategy
To solve an Integrated Concept Problem we must first identify the physical principles involved. Part (a) is related to the current topic. Part (b) involves a little math, primarily geometry. Part (c) requires an understanding of heat and density.
Solution to (a)
To a good approximation for a concave or semi-spherical surface, the point where the parallel rays from the sun converge will be at the focal point, so \(R=2 f=80.0 \mathrm{~cm}\).
Solution to (b)
The insolation is \(900 \mathrm{~W} / \mathrm{m}^{2}\). We must find the cross-sectional area \(A\) of the concave mirror, since the power delivered is \(900 \mathrm{~W} / \mathrm{m}^{2} \times \mathrm{A}\). The mirror in this case is a quarter-section of a cylinder, so the area for a length \(L\) of the mirror is \(\mathrm{A}=\frac{1}{4}(2 \pi R) \mathrm{L}\). The area for a length of 1.00 m is then
\[\mathrm{A}=\frac{\pi}{2} R(1.00 \mathrm{~m})=\frac{(3.14)}{2}(0.800 \mathrm{~m})(1.00 \mathrm{~m})=1.26 \mathrm{~m}^{2}. \nonumber\]
The insolation on the 1.00-m length of pipe is then
\[\left(9.00 \times 10^{2} \frac{\mathrm{W}}{\mathrm{m}^{2}}\right)\left(1.26 \mathrm{~m}^{2}\right)=1130 \mathrm{~W}. \nonumber\]
Solution to (c)
The increase in temperature is given by \(Q=m c \Delta T\). The mass \(m\) of the mineral oil in the one-meter section of pipe is
\[\begin{aligned}
m &=\rho \mathrm{V}=\rho \pi\left(\frac{d}{2}\right)^{2}(1.00 \mathrm{~m}) \\
&=\left(8.00 \times 10^{2} \mathrm{~kg} / \mathrm{m}^{3}\right)(3.14)(0.0100 \mathrm{~m})^{2}(1.00 \mathrm{~m}) \\
&=0.251 \mathrm{~kg}.
\end{aligned} \nonumber\]
Therefore, the increase in temperature in one minute is
\[\begin{aligned}
\Delta T &=Q / m c \\
&=\frac{(1130 \mathrm{~W})(60.0 \mathrm{~s})}{(0.251 \mathrm{~kg})\left(1670 \mathrm{~J} \cdot \mathrm{kg} /{ }^{\circ} \mathrm{C}\right)} \\
&=162^{\circ} \mathrm{C}.
\end{aligned} \nonumber\]
Discussion for (c)
An array of such pipes in the California desert can provide a thermal output of 250 MW on a sunny day, with fluids reaching temperatures as high as \(400^{\circ} \mathrm{C}\). We are considering only one meter of pipe here, and ignoring heat losses along the pipe.
What happens if an object is closer to a concave mirror than its focal length? This is analogous to a case 2 image for lenses (\(d_{\mathrm{o}}<f\) and \(f\) positive), which is a magnifier. In fact, this is how makeup mirrors act as magnifiers. Figure \(\PageIndex{6}\)(a) uses ray tracing to locate the image of an object placed close to a concave mirror. Rays from a common point on the object are reflected in such a manner that they appear to be coming from behind the mirror, meaning that the image is virtual and cannot be projected. As with a magnifying glass, the image is upright and larger than the object. This is a case 2 image for mirrors and is exactly analogous to that for lenses.
All three rays appear to originate from the same point after being reflected, locating the upright virtual image behind the mirror and showing it to be larger than the object. (b) Makeup mirrors are perhaps the most common use of a concave mirror to produce a larger, upright image.
A convex mirror is a diverging mirror (\(f\) is negative) and forms only one type of image. It is a case 3 image—one that is upright and smaller than the object, just as for diverging lenses. Figure \(\PageIndex{7}\)(a) uses ray tracing to illustrate the location and size of the case 3 image for mirrors. Since the image is behind the mirror, it cannot be projected and is thus a virtual image. It is also seen to be smaller than the object.
Example \(\PageIndex{3}\): Image in a Convex Mirror
A keratometer is a device used to measure the curvature of the cornea, particularly for fitting contact lenses. Light is reflected from the cornea, which acts like a convex mirror, and the keratometer measures the magnification of the image. The smaller the magnification, the smaller the radius of curvature of the cornea. If the light source is 12.0 cm from the cornea and the image’s magnification is 0.0320, what is the cornea’s radius of curvature?
Strategy
If we can find the focal length of the convex mirror formed by the cornea, we can find its radius of curvature (the radius of curvature is twice the focal length of a spherical mirror). We are given that the object distance is \(d_{\mathrm{o}}=12.0 \mathrm{~cm}\) and that \(m=0.0320\). We first solve for the image distance \(d_{\mathrm{i}}\), and then for \(f\).
Solution
\(m=-d_{\mathrm{i}} / d_{\mathrm{o}}\). Solving this expression for \(d_{\mathrm{i}}\) gives
\[d_{\mathrm{i}}=-m d_{\mathrm{o}}. \nonumber\]
Entering known values yields
\[d_{\mathrm{i}}=-(0.0320)(12.0 \mathrm{~cm})=-0.384 \mathrm{~cm}. \nonumber\]
\[\frac{1}{f}=\frac{1}{d_{\mathrm{o}}}+\frac{1}{d_{\mathrm{i}}} \nonumber\]
Substituting known values,
\[\frac{1}{f}=\frac{1}{12.0 \mathrm{~cm}}+\frac{1}{-0.384 \mathrm{~cm}}=\frac{-2.52}{\mathrm{~cm}}. \nonumber\]
This must be inverted to find \(f\):
\[f=\frac{\mathrm{cm}}{-2.52}=-0.400 \mathrm{~cm}. \nonumber\]
The radius of curvature is twice the focal length, so that
\[R=2|f|=0.800 \mathrm{~cm}. \nonumber\]
Discussion
Although the focal length \(f\) of a convex mirror is defined to be negative, we take the absolute value to give us a positive value for \(R\). The radius of curvature found here is reasonable for a cornea. The distance from cornea to retina in an adult eye is about 2.0 cm. In practice, many corneas are not spherical, complicating the job of fitting contact lenses. Note that the image distance here is negative, consistent with the fact that the image is behind the mirror, where it cannot be projected. In this section’s Problems and Exercises, you will show that for a fixed object distance, the smaller the radius of curvature, the smaller the magnification.
The three types of images formed by mirrors (cases 1, 2, and 3) are exactly analogous to those formed by lenses, as summarized in the table at the end of "Image Formation by Lenses". It is easiest to concentrate on only three types of images—then remember that concave mirrors act like convex lenses, whereas convex mirrors act like concave lenses.
TAKE-HOME EXPERIMENT: CONCAVE MIRRORS CLOSE TO HOME
Find a flashlight and identify the curved mirror used in it. Find another flashlight and shine the first flashlight onto the second one, which is turned off. Estimate the focal length of the mirror. You might try shining a flashlight on the curved mirror behind the headlight of a car, keeping the headlight switched off, and determine its focal length.
Section Summary
- The characteristics of an image formed by a flat mirror are: (a) The image and object are the same distance from the mirror, (b) The image is a virtual image, and (c) The image is situated behind the mirror.
-
Image length is half the radius of curvature.
\[f=\frac{R}{2} \nonumber\]
- A convex mirror is a diverging mirror and forms only one type of image, namely a virtual image.
Glossary
- converging mirror
- a concave mirror in which light rays that strike it parallel to its axis converge at one or more points along the axis
- diverging mirror
- a convex mirror in which light rays that strike it parallel to its axis bend away (diverge) from its axis
- law of reflection
- angle of reflection equals the angle of incidence


15.8B: Staphylococcal Food Poisoning
- Recognize the causes of staphylococcal food poisoning
Staphylococcus is a Gram-positive bacteria which includes several species that can cause a wide variety of infections in humans and other animals through infection or the production of toxins. Staphylococcal toxins are a common cause of food poisoning, as they can be produced in improperly-stored food. The main coagulase-positive staphylococcus is Staphylococcus aureus . These bacteria can survive on dry surfaces, increasing the chance of transmission.
Any S. aureus infection can cause the staphylococcal scalded skin syndrome, a cutaneous reaction to exotoxin absorbed into the bloodstream. It can also cause a type of septicaemia called pyaemia. The infection can be life-threatening. Problematically, Methicillin-resistant Staphylococcus aureus (MRSA) has become a major cause of hospital-acquired infections and is being recognized with increasing frequency in community-acquired infections.
Foodborne illness usually arises from improper handling, preparation, or food storage. Good hygiene practices before, during, and after food preparation can reduce the chances of contracting an illness. There is a consensus in the public health community that regular hand-washing is one of the most effective defenses against the spread of foodborne illness. The action of monitoring food to ensure that it will not cause foodborne illness is known as food safety.
Foodborne disease can also be caused by a large variety of toxins that affect the environment, such as pesticides or medicines in food, and naturally toxic substances such as poisonous mushrooms or reef fish. In the past, bacterial infections were thought to be more prevalent because few places had the capability to test for norovirus and no active surveillance was being done for this particular agent. Toxins for bacterial infections are delayed because the bacteria need time to multiply. Their symptoms are usually not seen until 12–72 hours or more after eating contaminated food.
Key Points
- Staphylococcus is a Gram-positive bacteria which includes several species that can cause a wide variety of infections in humans and other animals through infection or the production of toxins.
- Foodborne illness usually arises from improper handling, preparation, or storage of food.
- Good hygiene practices before, during, and after food preparation can reduce the chances of contracting an illness.
Key Terms
- toxin : A toxic or poisonous substance produced by the biological processes of biological organisms.
- norovirus : The genus of a number of species of virus, family Caliciviridae, causing human gastroenteritis, of which Norwalk virus is the prototype.


8.2.2: Racial and Ethnic Health Inequities
A recent report from the Office of Minority Health, an agency of the U.S. Department of Health and Human Services (DHHS), outlines how federal policies have caused and maintained structural disadvantages to obtaining long, healthy lives in minoritized communities. This report also uses the term minoritized groups rather than minority or minorities to further describe how structures and policies have affected populations, rather than these disparities being a result of numbers of a specific demographic in a population. (These terms will also be adopted in this text). The assertion is that racial and ethnic disparities in health and health outcomes are significantly caused or made worse by historical or existing governmental policies, and therefore a path toward health equity involves new federal policies and accountability (National Academies of Sciences, Engineering, and Medicine, 2023).
As discussed in Chapter 3 of this text, the United States’ early history of Native American extermination, land seizure, kidnapping of native children and required assimilation has caused, and continues to perpetuate significant health repercussions for American Indians and Alaskan Natives (AIAN). Concurrently, the forced immigration and enslavement of Africans, subsequent Jim Crow laws, and even purposeful public health abuses (such as the Tuskegee Syphilis Study), have negatively affected the health of Black Americans for centuries. Additionally, federal policies relating to immigration and social services also play a role in health disparities, particularly those experienced by immigrants and asylum seekers, and their children. Although health disparities are also correlated with poverty and lack of education, those racial and ethnic inequities exist at all socioeconomic strata - indicating that the explanation lies beyond income or education alone. A section in the report summarizes just some of the health inequities in America:
There are higher rates of childhood asthma among low-income households, higher morbidity and mortality from chronic diseases among individuals with lower educational attainment, and higher exposure to air pollution among residents of disinvested communities—disproportionately individuals who are racially and ethnically minoritized. Moreover, the effects of the structural determinants of health on many health outcomes persist when accounting for income and education (National Academies of Sciences, Engineering, and Medicine., 2023).
According to an earlier DHHS Health Equity Report in 2019-2020, significant racial disparities in health and social determinants of health have persisted across the decades. Although measures like life expectancy and educational status have improved for all Americans, there are still differences in these and other measures between racial groups. Racially minoritized groups have twice or higher the poverty rates as non-Hispanic Whites do (U.S. Department of Health and Human Services, Health Resources and Services Administration, Office of Health Equity., 2020). According to the most recent report from the U.S. Census Bureau tracking poverty rates for the year 2022, the poverty rate for those identifying as non-Hispanic White or Asian was 8.6%, whereas it was 17.1% and 16.9% for those identifying as Black and Hispanic respectively, and 25% for American Indian and Alaskan Natives (Shrider & Creamer, 2023).
This disparity in poverty rates is also reflected in the increased likelihood of minoritized populations living in impoverished zip codes. Southeastern and Southwestern states tend to have the highest poverty rates, a trend which has remained consistent over time, and is consistent with higher unemployment rates in these areas. Neighborhoods that have been historically racially segregated as “non-White” have associations with lower life expectancy and higher rates of homicide, infant mortality, and all-cause mortality, as well as higher rates of mental distress, community violence, excessive drinking and smoking, and HIV prevalence (U.S. Department of Health and Human Services, Health Resources and Services Administration, Office of Health Equity., 2020). These neighborhoods often also have higher exposures to environmental toxins like air pollution and lead (Doctrow, 2022).
According to the KFF (formerly known as the Kaiser Family Foundation) health tracking polling, many disparities exist with how safe minoritized groups feel in their homes and neighborhoods, and how discrimination impacts everything from housing to healthcare. This foundation conducted a survey on racism, discrimination, and health in a representative sample of 6,000 Americans in 2023. Hispanic, Black, Asian and AIAN individuals are less likely to report feeling safe in their neighborhoods. About twice as many racially minoritized respondents reported a family member being victimized by violence as did White respondents. Also consistent with earlier surveys, minoritized groups are significantly more likely to have a family member who has experienced mistreatment from law enforcement (Artiga, 2023). “Nearly half of Black Americans say they have been afraid their life was in danger due to their racial background,” (Reich, 2022). All of these disparities impact health on multiple levels across the lifespan; not the least of which is the heightened chronic stress due to discrimination, racism, and poverty.


1.20.5: Surfactants and Micelles: Mixed
In many industrial and commercial applications, mixed surfactant systems are used [1]. An extensive literature examines the properties of these systems [2-8].
A given aqueous solution contains two surfactants \(\mathrm{X}\) and \(\mathrm{Y}\) at temperature \(\mathrm{T}\) and pressure \(\mathrm{p}\). The critical micellar concentrations are \(\mathrm{cmc}_{\mathrm{X}}^{0}\) and \(\mathrm{cmc}_{\mathrm{Y}}^{0}\). For a solution containing both surfactants \(\mathrm{X}\) and \(\mathrm{Y}\), the critical micellar concentration of the mixed surfactant is cmc(mix). Here we use a pseudo-separate phase model for the micelles. The system under consideration comprises \(\mathrm{n}_{\mathrm{X}}^{0}\) and \(\mathrm{n}_{\mathrm{Y}}^{0}\) moles of the two surfactants. [Here the superscript ‘zero’ refers to the composition of the solution as prepared using the two pure surfactants.] A property \(\mathrm{r}\) is the ratio of the concentration of surfactant \(\mathrm{Y}\) to the total concentration of the two surfactants in the solution. Thus,
\[\mathrm{r}=\frac{\mathrm{n}_{\mathrm{Y}}^{0}}{\mathrm{n}_{\mathrm{X}}^{0}+\mathrm{n}_{\mathrm{Y}}^{0}}=\frac{\mathrm{c}_{\mathrm{Y}}^{0}}{\mathrm{c}_{\mathrm{X}}^{0}+\mathrm{c}_{\mathrm{Y}}^{0}} \nonumber \]
Here \(\mathrm{c}_{\mathrm{X}}^{0}\) and \(\mathrm{c}_{\mathrm{Y}}^{0}\) are the concentrations of the two surfactants in solution. We define a model system where cmc(mix) is a linear function of the property \(\mathrm{r}\); equation (b).
\[c m c(\operatorname{mix})=\left(\mathrm{cmc}_{\mathrm{Y}}^{0}-\mathrm{cmc}_{\mathrm{X}}^{0}\right) \, \mathrm{r}+\mathrm{cmc}_{\mathrm{X}}^{0} \nonumber \]
Or,
\[\mathrm{cmc}(\mathrm{mix})=\mathrm{r} \, \mathrm{cmc}_{\mathrm{Y}}^{0}+\mathrm{cmc}_{\mathrm{X}}^{0} \,(1-\mathrm{r}) \nonumber \]
Hence the critical micellar concentration of the two surfactants in a given solution, \(\mathrm{cmc}_{\mathrm{X}}\) and \(\mathrm{cmc}_{\mathrm{Y}}\), depend on parameter \(\mathrm{r}\).
\[\mathrm{cmc}_{\mathrm{Y}}=\mathrm{cmc}_{\mathrm{Y}}^{0} \, \mathrm{r} \nonumber \]
\[\mathrm{cmc}_{\mathrm{X}}=\mathrm{cmc}_{\mathrm{X}}^{0} \,(1-\mathrm{r}) \nonumber \]
Clearly in the absence of surfactant \(\mathrm{Y}\), micelles are not formed by surfactant \(\mathrm{X}\) until the concentration exceeds \(\mathrm{cmc}_{\mathrm{X}}^{0}\). If surfactant \(\mathrm{Y}\) is added to the solution, the \(\mathrm{cmc}_{\mathrm{X}}\) changes. In other words the properties of surfactants \(\mathrm{X}\) and \(\mathrm{Y}\) in a given solution are linked. We anticipate that for real system cmc(mix) is a function of \(\mathrm{c}_{\mathrm{X}}^{0}\) and \(\mathrm{c}_{\mathrm{Y}}^{0}\) so that cmc(mix) is a function of ratio \(\mathrm{r}\) and a quantity \(\theta\). The latter takes account of surfactant-surfactant interactions in the micellar pseudophase. Then
\[\mathrm{cmc}(\operatorname{mix})=\mathrm{r} \,\left(\mathrm{cmc}_{\mathrm{Y}}^{0}-\mathrm{cmc}_{\mathrm{X}}^{0}\right) \, \exp [-\theta \,(1-\mathrm{r})]+\mathrm{cmc}_{\mathrm{X}}^{0} \nonumber \]
For the surfactants \(\mathrm{X}\) and \(\mathrm{Y}\),
\[\mathrm{cmc}_{\mathrm{Y}}(\mathrm{mix})=\mathrm{r} \, \mathrm{cmc}_{\mathrm{Y}}^{0} \, \exp [-\theta \,(1-\mathrm{r})] \nonumber \]
and
\[\mathrm{cmc}_{\mathrm{x}}(\mathrm{mix})=\mathrm{cmc}_{\mathrm{x}}^{0} \,\{1-\mathrm{r} \, \exp [-\theta \,(1-\mathrm{r})]\} \nonumber \]
Therefore we envisage that the cmc of solutions containing a mixture of surfactants differs from that for model systems.
We turn attention to the enthalpies of mixed surfactant solutions. In the case of a mixed aqueous solution containing surfactants \(\mathrm{X}\) and \(\mathrm{Y}\), the partial molar enthalpies of the surfactants are anticipated to depend on their concentrations. We characterize a given system by a single enthalpic interaction parameter, \(\mathrm{h}(\text {int})\).
\[\mathrm{H}_{\mathrm{x}}(\mathrm{aq})=\mathrm{H}_{\mathrm{x}}^{\infty}(\mathrm{aq})+\mathrm{h}(\text {int}) \,\left(\mathrm{c}_{\mathrm{x}} / \mathrm{c}_{\mathrm{r}}\right) \nonumber \]
\[\mathrm{H}_{\mathrm{Y}}(\mathrm{aq})=\mathrm{H}_{\mathrm{Y}}^{\infty}(\mathrm{aq})+\mathrm{h}(\text { int }) \,\left(\mathrm{c}_{\mathrm{Y}} / \mathrm{c}_{\mathrm{r}}\right) \nonumber \]
Here \(\mathrm{H}_{\mathrm{x}}^{\infty}(\mathrm{aq})\) and \(\mathrm{H}_{\mathrm{Y}}^{\infty}(\mathrm{aq})\) are the ideal (infinite dilution) partial molar enthalpies of the two monomeric surfactants in aqueous solutions at defined \(\mathrm{T}\) and \(\mathrm{p}\).
The micellar pseudo-separate phase comprises two surfactants amounts \(\mathrm{n}_{X}(\text { mic })\) and \(\mathrm{n}_{Y}(\text { mic })\). The mole fractions \(\mathrm{x}_{\mathrm{X}}(\mathrm{mic})\) and \(\mathrm{x}_{\mathrm{X}}(\mathrm{mic}) \left[=\left(1-\mathrm{x}_{\mathrm{X}}(\mathrm{mic})\right]\) are given by equation (k).
\[\mathrm{x}_{\mathrm{X}}(\mathrm{mic})=\mathrm{n}_{\mathrm{X}}(\mathrm{mic}) /\left[\mathrm{n}_{\mathrm{x}}(\mathrm{mic})+\mathrm{n}_{\mathrm{Y}}(\mathrm{mic})\right]=1-\mathrm{x}_{\mathrm{Y}}(\mathrm{mic}) \nonumber \]
We relate the partial molar enthalpies \(\mathrm{H}_{\mathrm{X}}(\text {mic})\) and \(\mathrm{H}_{\mathrm{Y}}(\text {mic})\) in the mixed pseudo-separate phase to the molar enthalpies of surfactants \(\mathrm{X}\) and \(\mathrm{Y}\) in pure pseudo-separate micellar phases, \(\mathrm{H}_{\mathrm{X}}^{*}(\mathrm{mic})\) and \(\mathrm{H}_{\mathrm{Y}}^{*} \text { (mic) }\) using equations (l) and (m) where \(\mathrm{U}\) is a surfactant-surfactant interaction parameter.
\[\mathrm{H}_{\mathrm{X}}(\text { mic })=\mathrm{H}_{\mathrm{X}}^{*}(\mathrm{mic})+\left[1-\mathrm{x}_{\mathrm{X}}(\mathrm{mic})\right]^{2} \, \mathrm{U} \nonumber \]
\[\mathrm{H}_{\mathrm{Y}}(\mathrm{mic})=\mathrm{H}_{\mathrm{Y}}^{*}(\text { mic })+\left[\mathrm{x}_{\mathrm{X}}(\mathrm{mic})\right]^{2} \, \mathrm{U} \nonumber \]
Footnotes
[1] J. H. Clint, J. Chem. Soc. Faraday Trans.,1, 1975, 71 ,1327.
[2] P. M .Holland, Adv. Colloid Interface Sci.,1986, 26 ,111; and references therein.
[3] A. H. Roux, D. Hetu, G. Perron and J. E. Desnoyers, J. Solution Chem.,1984, 13 ,1.
[4] M. J. Hey, J. W. MacTaggart and C. H. Rochester, J. Chem. Soc. Faraday Trans.1, 1985, 81 ,207.
[5] J. L. Lopez-Fontan, M. J. Suarez, V. Mosquera and F. Sarmiento, Phys. Chem. Chem. Phys.,1999, 1 ,3583.
[6] R. DeLisi, A. Inglese, S. Milioto and A. Pellerito, Langmuir, 1997, 13 ,192.
[7] M. J. Blandamer, B. Briggs, P. M. Cullis and J. B. F. N. Engberts, Phys. Chem. Chem. Phys.,2000, 2 ,5146.
[8] J. F. Rathman and J. F. Scamehorn, Langmuir, 1988, 4 ,474.
[9] A. H. Roux, D. Hetu, G. Perron and J. E. Desn
oyers, J. Solution Chem.,1984, 13 ,1.


3.4: Finite and infinite relations
Sets can be infinite, and relations can be too. An infinite relation is simply a relation with infinitely many ordered pairs in it. This might seem strange at first, since how could we ever hope to specify all the ordered pairs? But it’s really no different than with sets: we either have to do it intensionally, or else have a rule for systematically computing the extension.
As an example of the first, consider the relation “isGreaterThan" between \(\mathbb{Z}\) and \(\mathbb{Z}\) . (Recall that “ \(\mathbb{Z}\) " is just a way of writing “the set of integers.") This relation contains ordered pairs like (5, 2) and (17, –13), since 5 isGreaterThan 2 and 17 isGreaterThan –13, but not (7, 9) or (11, 11). Clearly it’s an infinite relation. We couldn’t list all the pairs, but we don’t need to, since the name implies the underlying meaning of the relation.
As an example of the second, consider the relation “isLuckierThan" between \(\mathbb{N}\) and \(\mathbb{N}\) . (The “ \(\mathbb{N}\) " means “the natural numbers.") We specify it extensionally as follows:
{ (1, 13), (2, 13), (3, 13), …(12, 13), (14, 13), (15, 13), (16, 13), … }
Here we’re just saying “every number is luckier than 13 (except for 13 itself, of course)."


Chapter 10
Liquids and solids are similar in that they are matter composed of atoms, ions, or molecules. They are incompressible and have similar densities that are both much larger than those of gases. They are different in that liquids have no fixed shape, and solids are rigid.
They are similar in that the atoms or molecules are free to move from one position to another. They differ in that the particles of a liquid are confined to the shape of the vessel in which they are placed. In contrast, a gas will expand without limit to fill the space into which it is placed.
All atoms and molecules will condense into a liquid or solid in which the attractive forces exceed the kinetic energy of the molecules, at sufficiently low temperature.
(a) Dispersion forces occur as an atom develops a temporary dipole moment when its electrons are distributed asymmetrically about the nucleus. This structure is more prevalent in large atoms such as argon or radon. A second atom can then be distorted by the appearance of the dipole in the first atom. The electrons of the second atom are attracted toward the positive end of the first atom, which sets up a dipole in the second atom. The net result is rapidly fluctuating, temporary dipoles that attract one another (e.g., Ar). (b) A dipole-dipole attraction is a force that results from an electrostatic attraction of the positive end of one polar molecule for the negative end of another polar molecule (e.g., ICI molecules attract one another by dipole-dipole interaction). (c) Hydrogen bonds form whenever a hydrogen atom is bonded to one of the more electronegative atoms, such as a fluorine, oxygen, or nitrogen atom. The electrostatic attraction between the partially positive hydrogen atom in one molecule and the partially negative atom in another molecule gives rise to a strong dipole-dipole interaction called a hydrogen bond (e.g.,
The London forces typically increase as the number of electrons increase.
(a) SiH 4 < HCl < H 2 O; (b) F 2 < Cl 2 < Br 2 ; (c) CH 4 < C 2 H 6 < C 3 H 8 ; (d) N 2 < O 2 < NO
Only rather small dipole-dipole interactions from C-H bonds are available to hold n -butane in the liquid state. Chloroethane, however, has rather large dipole interactions because of the Cl-C bond; the interaction, therefore, is stronger, leading to a higher boiling point.
−85 °C. Water has stronger hydrogen bonds, so it melts at a higher temperature.
The hydrogen bond between two hydrogen fluoride molecules is stronger than that between two water molecules because the electronegativity of F is greater than that of O. Consequently, the partial negative charge on F is greater than that on O. The hydrogen bond between the partially positive H and the larger partially negative F will be stronger than that formed between H and O.
H-bonding is the principle IMF holding the protein strands together. The H-bonding is between the and
(a) hydrogen bonding, dipole-dipole attraction, and dispersion forces; (b) dispersion forces; (c) dipole-dipole attraction and dispersion forces
The water molecules have strong intermolecular forces of hydrogen bonding. The water molecules are thus attracted strongly to one another and exhibit a relatively large surface tension, forming a type of “skin” at its surface. This skin can support a bug or paper clip if gently placed on the water.
Temperature has an effect on intermolecular forces: The higher the temperature, the greater the kinetic energies of the molecules and the greater the extent to which their intermolecular forces are overcome, and so the more fluid (less viscous) the liquid. The lower the temperature, the less the intermolecular forces are overcome, and so the less viscous the liquid.
(a) As the water reaches higher temperatures, the increased kinetic energies of its molecules are more effective in overcoming hydrogen bonding, and so its surface tension decreases. Surface tension and intermolecular forces are directly related. (b) The same trend in viscosity is seen as in surface tension, and for the same reason.
1.7 10 −4 m
The heat is absorbed by the ice, providing the energy required to partially overcome intermolecular attractive forces in the solid and causing a phase transition to liquid water. The solution remains at 0 °C until all the ice is melted. Only the amount of water existing as ice changes until the ice disappears. Then the temperature of the water can rise.
We can see the amount of liquid in an open container decrease and we can smell the vapor of some liquids.
The vapor pressure of a liquid decreases as the strength of its intermolecular forces increases.
As the temperature increases, the average kinetic energy of the molecules of gasoline increases and so a greater fraction of molecules have sufficient energy to escape from the liquid than at lower temperatures.
They are equal when the pressure of gas above the liquid is exactly 1 atm.
approximately 95 °C
(a) At 5000 feet, the atmospheric pressure is lower than at sea level, and water will therefore boil at a lower temperature. This lower temperature will cause the physical and chemical changes involved in cooking the egg to proceed more slowly, and a longer time is required to fully cook the egg. (b) As long as the air surrounding the body contains less water vapor than the maximum that air can hold at that temperature, perspiration will evaporate, thereby cooling the body by removing the heat of vaporization required to vaporize the water.
Dispersion forces increase with molecular mass or size. As the number of atoms composing the molecules in this homologous series increases, so does the extent of intermolecular attraction via dispersion forces and, consequently, the energy required to overcome these forces and vaporize the liquids.
The boiling point of CS 2 is higher than that of CO 2 partially because of the higher molecular weight of CS 2 ; consequently, the attractive forces are stronger in CS 2 . It would be expected, therefore, that the heat of vaporization would be greater than that of 9.8 kJ/mol for CO 2 . A value of 28 kJ/mol would seem reasonable. A value of −8.4 kJ/mol would indicate a release of energy upon vaporization, which is clearly implausible.
The thermal energy (heat) needed to evaporate the liquid is removed from the skin.
1125 kJ
(a) 13.0 kJ; (b) It is likely that the heat of vaporization will have a larger magnitude since in the case of vaporization the intermolecular interactions have to be completely overcome, while melting weakens or destroys only some of them.
At low pressures and 0.005 °C, the water is a gas. As the pressure increases to 4.6 torr, the water becomes a solid; as the pressure increases still more, it becomes a liquid. At 40 °C, water at low pressure is a vapor; at pressures higher than about 75 torr, it converts into a liquid. At −40 °C, water goes from a gas to a solid as the pressure increases above very low values.
(a) gas; (b) gas; (c) gas; (d) gas; (e) solid; (f) gas
Yes, ice will sublime, although it may take it several days. Ice has a small vapor pressure, and some ice molecules form gas and escape from the ice crystals. As time passes, more and more solid converts to gas until eventually the clothes are dry.
(a)
(b)
(c)
(d)
(e) liquid phase (f) sublimation
(e) molecular crystals
Ice has a crystalline structure stabilized by hydrogen bonding. These intermolecular forces are of comparable strength and thus require the same amount of energy to overcome. As a result, ice melts at a single temperature and not over a range of temperatures. The various, very large molecules that compose butter experience varied van der Waals attractions of various strengths that are overcome at various temperatures, and so the melting process occurs over a wide temperature range.
(a) ionic; (b) covalent network; (c) molecular; (d) metallic; (e) covalent network; (f) molecular; (g) molecular; (h) ionic; (i) ionic
X = ionic; Y = metallic; Z = covalent network
(b) metallic solid
The structure of this low-temperature form of iron (below 910 °C) is body-centered cubic. There is one-eighth atom at each of the eight corners of the cube and one atom in the center of the cube.
eight
12
(a) 1.370 Å; (b) 19.26 g/cm
(a) 2.176 Å; (b) 3.595 g/cm 3
The crystal structure of Si shows that it is less tightly packed (coordination number 4) in the solid than Al (coordination number 12).
In a closest-packed array, two tetrahedral holes exist for each anion. If only half the tetrahedral holes are occupied, the numbers of anions and cations are equal. The formula for cadmium sulfide is CdS.
Co 3 O 4
In a simple cubic array, only one cubic hole can be occupied be a cation for each anion in the array. The ratio of thallium to iodide must be 1:1; therefore, the formula for thallium is TlI.
59.95%; The oxidation number of titanium is +4.
Both ions are close in size: Mg, 0.65; Li, 0.60. This similarity allows the two to interchange rather easily. The difference in charge is generally compensated by the switch of Si 4+ for Al 3+ .
Mn 2 O 3
1.48 Å
2.874 Å
20.2°
1.74 10 4 eV


15.5: Life in the Universe (Exercises)
For Further Exploration
Articles
Astrobiology
Chyba, C. “The New Search for Life in the Universe.” Astronomy (May 2010): 34. An overview of astrobiology and the search for life out there in general, with a brief discussion of the search for intelligence.
Dorminey, B. “A New Way to Search for Life in Space.” Astronomy (June 2014): 44. Finding evidence of photosynthesis on other worlds.
McKay, C., & Garcia, V. “How to Search for Life on Mars.” Scientific American (June 2014): 44–49. Experiments future probes could perform.
Reed, N. “Why We Haven’t Found Another Earth Yet.” Astronomy (February 2016): 25. On the search for smaller earthlike planets in their star’s habitable zones, and where we stand.
Shapiro, R. “A Simpler Origin of Life.” Scientific American (June 2007): 46. New ideas about what kind of molecules formed first so life could begin.
Simpson, S. “Questioning the Oldest Signs of Life.” Scientific American (April 2003): 70. On the difficulty of interpreting biosignatures in rocks and the implications for the search for life on other worlds.
SETI
Chandler, D. “The New Search for Alien Intelligence.” Astronomy (September 2013): 28. Review of various ways of finding other civilizations out there, not just radio wave searches.
Crawford, I. “Where Are They?” Scientific American (July 2000): 38. On the Fermi paradox and its resolutions, and on galactic colonization models.
Folger, T. “Contact: The Day After.” Scientific American (January 2011): 40–45. Journalist reports on efforts to prepare for ET signals; protocols and plans for interpreting messages; and discussions of active SETI.
Kuhn, J., et al. “How to Find ET with Infrared Light.” Astronomy (June 2013): 30. On tracking alien civilizations by the heat they put out.
Lubick, N. “An Ear to the Stars.” Scientific American (November 2002): 42. Profile of SETI researcher Jill Tarter.
Nadis, S. “How Many Civilizations Lurk in the Cosmos?” Astronomy (April 2010): 24. New estimates for the terms in the Drake equation.
Shostak, S. “Closing in on E.T.” Sky & Telescope (November 2010): 22. Nice summary of current and proposed efforts to search for intelligent life out there.
Websites
Astrobiology
Astrobiology Web: http://astrobiology.com/ . A news site with good information and lots of material.
Exploring Life’s Origins: http://exploringorigins.org/index.html . A website for the Exploring Origins Project, part of the multimedia exhibit of the Boston Museum of Science. Explore the origin of life on Earth with an interactive timeline, gain a deeper knowledge of the role of RNA, “build” a cell, and explore links to learn more about astrobiology and other related information.
History of Astrobiology: https://astrobiology.nasa.gov/about/...-astrobiology/ . By Marc Kaufman, on the NASA Astrobiology site.
Life, Here and Beyond: https://astrobiology.nasa.gov/about/ . By Marc Kaufman, on the NASA Astrobiology site.
SETI
Berkeley SETI Research Center: https://seti.berkeley.edu/ . The University of California group has received a $100 million grant from a Russian-American billionaire to begin the Breakthrough: Listen project, a major step forward in the number of stars and number of radio channels being searched.
Planetary Society: www.planetary.org/explore/projects/seti/. This advocacy group for exploration has several pages devoted to the search for life.
SETI Institute: http://www.seti.org . A key organization in the search for life in the universe; the institute’s website is full of information and videos about both astrobiology and SETI.
SETI: http://www.skyandtelescope.com/tag/seti/ . Sky & Telescope magazine offers good articles on this topic.
Videos
Astrobiology
Copernicus Complex: Are We Special in the Cosmos?: https://www.youtube.com/watch?v=ERp0AHYRm_Q . A video of a popular-level talk by Caleb Scharf of Columbia University (1:18:54).
Life at the Edge: Life in Extreme Environments on Earth and the Search for Life in the Universe: https://www.youtube.com/watch?v=91JQmTn0SF0 . A video of a 2009 nontechnical lecture by Lynn Rothschild of NASA Ames Research Center (1:31:21).
Saturn’s Moon Titan: A World with Rivers, Lakes, and Possibly Even Life: https://www.youtube.com/watch?v=bbkTJeHoOKY . A video of a 2011 talk by Chris McKay of NASA Ames Research Center (1:23:33).
SETI
Allen Telescope Array: The Newest Pitchfork for Exploring the Cosmic Haystack: https://www.youtube.com/watch?v=aqsI1HZCgUM . A 2013 popular-level lecture by Jill Tarter of the SETI Institute (1:45:55).
Confessions of an Alien Hunter: fora.tv/2009/03/31/Seth_Shost...n_Alien_Hunter. 2009 interview with Seth Shostak on FORA TV (36:27).
Search for Extra-Terrestrial Intelligence: Necessarily a Long-Term Strategy: http://www.longnow.org/seminars/0200...term-strategy/ . 2004 talk by Jill Tarter at the Long Now Foundation (1:21:13).
Search for Intelligent Life Among the Stars: New Strategies: https://www.youtube.com/watch?v=m9WxW2ktcKU . A 2010 nontechnical talk by Seth Shostak of the SETI Institute (1:29:58).
Review Questions
- What is the Copernican principle? Make a list of scientific discoveries that confirm it.
- Where in the solar system (and beyond) have scientists found evidence of organic molecules?
- Give a short history of the atoms that are now in your little finger, going back to the beginning of the universe.
- What is a biomarker? Give some possible examples of biomarkers we might look for beyond the solar system.
- Why are Mars and Europa the top targets for the study of astrobiology?
- Why is traveling between the stars (by creatures like us) difficult?
- What are the advantages to using radio waves for communication between civilizations that live around different stars? List as many as you can.
- What is the “cosmic haystack problem”? List as many of its components as you can think of.
- What is a habitable zone?
- Why is the simultaneous detection of methane and oxygen in an atmosphere a good indication of the existence of a biosphere on that planet?
- What are two characteristic properties of life that distinguish it from nonliving things?
- What are the three requirements that scientists believe an environment needs to supply life with in order to be considered habitable?
- Can you name five environmental conditions that, in their extremes, microbial life has been challenged by and has learned to survive on Earth?
Thought Questions
- Would a human have been possible during the first generation of stars that formed right after the Big Bang? Why or why not?
- If we do find life on Mars, what might be some ways to check whether it formed separately from Earth life, or whether exchanges of material between the two planets meant that the two forms of life have a common origin?
- What kind of evidence do you think would convince astronomers that an extraterrestrial spacecraft has landed on Earth?
- What are some reasons that more advanced civilizations might want to send out messages to other star systems?
- What are some answers to the Fermi paradox? Can you think of some that are not discussed in this chapter?
- Why is there so little evidence of Earth’s earliest history and therefore the period when life first began on our planet?
- Why was the development of photosynthesis a major milestone in the evolution of life?
- Does all life on Earth require sunshine?
- Why is life unlikely to be found on the surface of Mars today?
- In this chapter, we identify these characteristic properties of life: life extracts energy from its environment, and has a means of encoding and replicating information in order to make faithful copies of itself. Does this definition fully capture what we think of as “life”? How might our definition be biased by our terrestrial environment?
- Given that no sunlight can penetrate Europa’s ice shell, what would be the type of energy that could make some form of europan life possible?

- In addition to an atmosphere dominated by nitrogen, how else is Saturn’s moon Titan similar to Earth?
- How can a planet’s atmosphere affect the width of the habitable zone in its planetary system?
- Why are we limited to finding life on planets orbiting other stars to situations where the biosphere has created planet-scale changes?


8: Memos
Chapter Objectives
The purpose of this chapter is to:
- Describe the purpose of a memo
- Identify the proper format of a memo
- Explain how to effectively organize a memo's content
Once the dominant form of communication in the workplace, memos typically serve as a tool for internal communication within organization. Memos can update policies and procedures, announce meetings or organizational changes, or inform the internal audience as needed. The truth is, there are lots of different occasions to write memos. We use memos to distribute progress reports, meeting minutes, incident reports, directive, and responses to an inquiry.
Memos are an accepted method of communication within a specific business, company, or institution. The fact that memos are only used internally makes them different from letters, emails, and texting, which can be used inside and outside of the workplace. The successful operation of a company depends on memos for communication between the employees of the company. They are often (but not always) written from a one-to-all perspective, broadcasting a message to an audience, rather than a one-on-one, interpersonal communication. Memos may be used to update a team on activities for a given project or to inform a specific group within a company of an event, action, or observance.
Memo Purpose
A memo’s purpose is often to inform, but it may occasionally include an element of persuasion or a call-to-action. All organizations have informal and formal communication networks. The unofficial, informal communication network within an organization is often referred to as the grapevine, and it is characterized by gossip, rumors, and innuendo. On the grapevine, one person may hear that someone else is going to be laid off and start passing the news around. Rumors change and transform as they are passed from person to person, and before you know it, the word is that they are shutting down your entire department!
One effective way to address unofficial speculation is to spell out clearly for all employees what is going on with a particular issue. If budget cuts are a concern, then you could send a memo explaining the changes that are imminent. If a company wants employees to take action, they may also issue a memo. For example, on February 13, 2009, upper management at the Panasonic Corporation issued a declaration that all employees should buy at least $1,600 worth of Panasonic products. The company president noted that if everyone supported the company with purchases, it would benefit all (Lewis, 2009). While memos do not normally include a call-to-action that requires personal spending, they do usually represent the organization’s interests. They may also include statements that align business and employee interests.
Memo Format
Memos are typically brief, concise, organized for readability, and addressed to targeted audiences with specific subject lines. A memo’s format provides employees with clear and easy access to information. The message is direct. In order to be easy to navigate, a memo follows a specific format. Usually, we start by typing the word Memo or Memorandum under the letterhead. (Note that there are memo templates available to you in Microsoft Word.) If there is no letterhead, start by typing up the name of the company on top of the page before you write the word Memo or Memorandum. Then, follow with a heading block.
Organization of a Heading Block
To: Date:
From: OR To:
Date: From:
Subject: Subject:
The recipient can be a single person or a group of people. If you are writing the memo to one person, use the person’s name and job title. If you are writing to a group of people, use the name of the group rather than a list of specific names. For example, we can address a memo to The Sales Team. The sender should always include their name and their job title. The subject line is very important and should be used responsibly. It must be brief and concise while specifically indicating the main point of the memo. Note that every major word should start with a capital letter (prepositions and articles are not capitalized in subject lines and titles in general). Some companies prefer that the subject line is also in bold.
Example of a Heading Block
To: Student Support Services
From: Greg Brown, User Support Technician
Date: December 1, 2020
Subject: Server Outage on Friday, December 4
You will fill in the information, and then you will begin your memo. If it is a longer memo (longer than a few paragraphs), you may also wish to use section headers. Remember that we do not start the memo with any type of salutation (such as, "Dear Assistant Managers"). Also note that the order of the parts of the memo heading block varies according to different companies’ expectations.
Unlike emails and letters, memos do not have a closing and signature. Instead, in the past, the writer would indicate that they really wrote the memo by initialing by their name in the paper heading using blue or black ink. Or, if a memo was typed up by another person, it would be taken back to a supervisor for approval, and the person who typed the memo would initial at the bottom of the memo like this, for example: /jc. However, in today's workplace, most employees write their own messages on their own computers, so this is rarely done anymore.
Finally, in the days of paper, memos also indicated attachments. If a memo arrived on a desk with the notation "Enclosure" it meant that there was supposed to be something paper clipped to the memo with additional information. And, if the recipient saw nothing paper clipped, it was immediately clear that something was missing. It is for this reason that you often see in an email a paper clip icon for the attachment button or notification. Figure 1 shows an example of a paper memorandum.
Figure 1: Memorandum on Attire for Johnson/Robb wedding from 15 November 1967 was sent to the Commandant of the Marine Corps, Wallace Greene, Jr. in regards to the appropriate attire to wear to the Lynda Johnson/Charles Robb wedding at the White House. From the Charles Robb Collection (COLL/591), Marine Corps Archives & Special Collections, Quantico, Virginia. "Memorandum on Attire for Johnson/Robb Wedding, 15 November 1967" by Archives Branch, USMC History Division is licensed under CC BY 2.0
Memo Content
In a standard writing format, we might expect to see an introduction , a body , and a conclusion . All these are present in a memo, and each part has a purpose. The introduction in the opening uses a declarative sentence to announce the main topic. If there are any effective dates or due dates that need to be presented, they should be presented in the first paragraph to ensure that they are not overlooked.
The body elaborates or lists major points associated with the topic, and the conclusion serves as a summary. It is quite common at the end of the memo to encourage readers to follow up with questions or comments and to provide contact information for their convenience. Throughout the memo, we need to pay attention to the tone to maintain goodwill.
Example of Memo Format and Content
To: All Employees
From: Maya James, President, Blue Lake Community College
Date: September 21, 2020
Subject: Future Expenditure Guidelines
After careful deliberation, I have determined it is necessary to begin the initial steps of a financial stewardship program that carries Blue Lake Community College through what appears to be a two-year cycle of a severe shortfall in revenue and subsequent necessary legislative budget reductions.
Beginning October 1, 2020, the following actions are being implemented for the General Fund, Auxiliary Fund, and Capital Fund in order to address the projected reductions in funding for the remainder of this year, 2020/2021, and for the next year, 2021/2022:
- Only purchases needed to operate the university should be made so that we can begin saving to reduce the impact of the 2021/2022 budget reductions.
- Requests for out-of-state travel will require approval from the Executive Committee to ensure that only necessary institutional travel occurs.
- Purchase, including in-state travel and budget transfers, will require the appropriate vice president's approval.
Please understand that we are taking these prudent steps to create savings that will allow BLCC to reduce the impact of projected cuts in expected 2021/2022 legislative reductions. Thank you for your cooperation. Please direct any questions to my office at<PHONE_NUMBER> or email at<EMAIL_ADDRESS>.
General Business Memo Tips
- Audience Orientation - Always consider the audience and their needs when preparing a memo. An acronym or abbreviation that is known to management may not be know by all the employees of the organization, so if the memo is to be posted and distributed within the organization, your goal should be clear and concise communication at all levels with no ambiguity.
- Professional Tone - Memos are often announcements, and the person sending the memo speaks for a part or all of the organization. While it may contain a request for feedback, the announcement itself is linear, from the organization to the employees. The memo may have legal standing, as it often reflects policies and procedures.
- Subject Emphasis - The subject is normally declared in the subject line and should be clear and concise. If the memo is announcing the observance of a holiday, for example, the specific holiday should be named in the subject line--for example, use "Thanksgiving Weekend Schedule" rather than "Holiday Observance."
- Direct Format - Memos are always direct, meaning they get to the point quickly and the purpose is clearly announced.
- Objectivity - Memos should contain just facts and should have an objective tone without personal bias, preference, or interest on display. Avoid subjectivity.
All links live as of June 2021.
This work "Writing Effective Memos" is a derivative of " ENGL 145: Technical and Report Writing " by Amber Kinonen, “ Open Technical Communication ” by Tiffani Reardon, Tamara Powell, Jonathan Arnett, Monique Logan, and Cassandra Race, and “ Professional Communications ” by JR Dingwall, Chuck Labrie, TK McLennon, and Laura Underwood, used under a CC BY license. "Writing Effective Memos" is licensed under CC BY 4.0 by Iva Balic.


2.22: Gramática- El verbo estar + preposiciones / ¿Dónde?
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- Erica Brown, Alejandra Escudero, María Cristina Montoya, & Elizabeth Small
- SUNY Oneonta via OER SUNY
Objetivos
- Recognize the correct conjugations of the verb estar
- Use the prepositions of location to describe where things and people are located in a house
Estar
You have already learned the verb ser, and you’ve used it to: introduce yourself and others, to describe physical characteristics and personality traits, to indicate place of origin or nationality, to tell time, and to give dates. Spanish has another verb that also means “to be” – estar . You used estar at the beginning of this course to indicate how you are feeling (¿Cómo estás? Estoy bien, gracias). In addition to indicating how you and others are feeling (which we’ll discuss in a future chapter), estar is used to express location. So, if you are talking about where a person or a thing is located, you will use estar :
| estar (to be) | ||
| singular | plural | |
| 1 a |
yo estoy |
nosotros estamos |
| 2 a |
tú estás |
vosotros estáis |
| 3 a |
él ella usted está |
ellos ellas ustedes están |
As you should notice, e star has an irregular yo form along with accented endings in all but the nosotros form.
Estar y la localización
To use estar to express the locations of people and things, you’ll need to know some prepositions of location as well:
- alrededor ( around )
- al lado de ( beside, next to )
- enfrente de ( in front of )
- a la derecha de ( to the right of )
- a la izquierda de ( to the left of )
- debajo de ( under )
- en ( on, in )
- encima de ( on, above )
- detrás de ( behind )
- cerca de ( close to )
- lejos de ( far from )
- Alegría está en su dormitorio. (Alegría is in her bedroom.)
- ¿Dónde está tu restaurante favorito? (Where is your favorite restaurant?)
- Los padres de Benjamín están de vacaciones en Cancún. (Ben’s parents are in Cancún on vacation.)
- Ay de mí, ¿dónde están mis llaves? (Good grief, where are my keys?)
- La cafetería está enfrente de la librería. (The cafeteria is in front of the bookstore.)
- No estoy en clase ahora porque estoy enfermo. (I’m not in class right now because I’m sick.)
- Nosotros estamos cerca del museo pero no podemos encontrarlo. (We’re close to the museum, but we can’t find it.)
¿Dónde?
The question word used to ask about locations is ¿Dónde? (where?). It can be combined with prepositions to ask more specific questions, for example: ¿De dónde? (from where?) and ¿Adónde? (to where?). Remember that question words always need an accent mark, and don’t forget to put the upside-down question mark at the beginning of the question with the upside-right question mark at the end of the question.
- ¿ De dónde eres? ( Where are you from? )
- Soy de Oneonta, Nueva York. ( I am from Oneonta, New York. )
- Ah, y ¿ dónde está Oneonta? ( Ah, and where is Oneonta? )
- Oneonta está en las montañas; no está cerca de la ciudad de Nueva York. ( Oneonta is in the mountains; it’s not close to New York City. )
"Estar" with prepositions and adjectives
In 1.4.0 you learned how to conjugate the verb "ser" ( to be ). Unlike English, Spanish has a second verb that translates as "to be": "estar".
Conjugation
While it looks like a regular "-ar" verb, "estar" is an irregular verb because the "yo" form is irregular; and the endings of " tú, usted, él, ella, ustedes, ellos, ellas" require a written accent.
| Singular | Conjugation | Plural | Conjugation |
|---|---|---|---|
| yo | estoy | nosotros/nosotras | estamos |
| tú | estás | vosotros/vosotras | estáis |
| usted, él/ella | está | ustedes, ellos/ellas | están |
Uses of "estar"
The acronym used to help you remember the uses of "estar" is PLACE:
In the following list, the words that refer to the acronym PLACE are indicated in bold .
- Position: "El libro está encima de la mesa." "El lápiz está en la mochila." ("The book is on top of the table." "The pencil is in the backpack.")
- Location: "Ramiro está en la oficina." "Inés está en casa." ("Ramiro is at the office." "Inés is at home.")
- Action: "Ellos están hablando". "Beto está caminando." ("They are talking." "Beto is walking.")
- Condition: "Estoy enfermo." "Laura está preocupada." ("I am sick (ill)." "Laura is worried.")
- Emotion: "Tú estás feliz." "Vosotros estáis tristes." ("You are happy." "You are sad.")
Common prepositions and adverbs used with location are:
| al lado de | next to | delante de | in front of (not facing) |
| a la derecha de | to the right of | detrás de | behind |
| a la izquierda de | to the left of | en | in; on; at |
| allá | over there | encima de | on top of |
| allí | there | en frente de | in front of (facing) |
| aquí | here | entre | between |
| cerca de | near | lejos de | far from |
| con | with | sin | without |
| debajo de | below | sobre | on; over |
Common adjectives of condition and emotion are:
| abierto/a | open | contento/a | content | listo/a | ready |
| aburrido/a | bored | desordenado/a | messy; disorderly | nervioso/a | nervous |
| alegre | happy | enamorado/a (de) | in love (with) | ocupado/a | busy |
| avergonzado/a | embarrassed | enojado/a | angry | ordenado/a | tidy; orderly |
| cansado/a | tired | enfermo/a | sick (ill) | preocupado/a (por) | worried (about) |
| cerrado/a | closed | equivocado/a | wrong | seguro/a | sure |
| cómodo/a | comfortable | feliz | happy | sucio/a | dirty |
| confundido/a | confused | limpio/a | clean | triste | sad |
A Practicar
Contributors and Attributions
- Gramu00e1tica: El verbo estar + preposiciones. Authored by : SUNY Oneonta with Lumen Learning. Provided by : SUNY Oneonta. License : CC BY: Attribution