<s>[INST] Find $\sin \left(\frac x2\right)$ and $\cos \left(\frac x2\right)$ given $\cos(x) =\frac{-15}{17}$ and $x$ in quadrant II I'm having trouble fully understanding these two problems. I know I have to use the half angle formula for one, and the double for another. The first question goes as this 
Given that  $$\cos(x) =\frac{-15}{17}$$ and $x$ in quadrant II find the exact values of: 
$$\sin \left(\frac x2\right)$$ and $$\cos \left(\frac x2\right)$$
For this problem the biggest question I do have is what should I input for $x$ exactly? It states $\cos(x)$ is a fraction, so should I simply add that as $x$ in $$\frac x2 $$? Or if I need to find what $$\sin\left(\frac x2\right)$$ is how could I find that? 
Secondly, what would the best course of action be to see if it would be a negative or positive since in QII it could be either or.
Second is Given $$\sin(x) = \frac{-7}{25}$$ and $x$ in quadrant III, find the value of $\sin(2x)$,$\cos(2x)$,$\tan(2x)$.
This one I am kinda lost, would I need to find the value of what sin is and just plug it into what x is (For each corresponding formula.)? Or tackle it differently?
This is somewhat a loaded question, and I am sorry for that. But I do appreciate any help anyone could give. 
 [/INST] I always forget the formulas for stuff like $\sin\left(\frac{x}{2}\right)$. But you don't need to remember them. You only need to know two important formulas. 
\begin{align}
1 = \cos^2(x) + \sin^2(x)  \tag{1} \\ \\
\cos(2x) = \cos(x + x) = \cos(x)\cos(x) - \sin(x)\sin(x) = &\cos^2(x) - \sin^2(x) \tag{2}
\end{align}
so we end up with 
\begin{align}
1 &= \cos^2(x) + \sin^2(x)  \tag{1} \\ \\
\cos(2x) & = \cos^2(x) - \sin^2(x) \tag{2}
\end{align}
But in these equations we didn't have to use $x$, we could have used $y$ or $a$ or $5$. They are all true no matter what real value we stick in there. So watch what happens if instead of putting $x$ in there we put $x/2$. 
\begin{align}
1 &= \cos^2\left(\frac{x}{2}\right) +  \sin^2\left(\frac{x}{2}\right) \tag{1} \\ \\
\cos\left(2\frac{x}{2}\right) & = \cos^2\left(\frac{x}{2}\right) - \sin^2\left(\frac{x}{2}\right) \tag{2}
\end{align}
So here we are
\begin{align}
1 &= \cos^2\left(\frac{x}{2}\right) +  \sin^2\left(\frac{x}{2}\right)  \tag{1} \\ \\
\cos\left(x\right) & = \cos^2\left(\frac{x}{2}\right) - \sin^2\left(\frac{x}{2}\right) \tag{2}
\end{align}
To get the half angle formulas all we have to do is add or substract these two equations. For instance, if we subtract $(2)$ from $(1)$ we get
\begin{align}
1 - \cos(x) = 2\sin^2\left(\frac{x}{2}\right) \tag{3}
\end{align}
Now, I'm going to plug in what you said that $\cos(x) = \frac{-15}{17}$ into $(3)$ just above: 
\begin{align}
1 - \frac{-15}{17} &= 2\sin^2\left(\frac{x}{2}\right) \\ \\
1 + \frac{15}{17} & = 2\sin^2\left(\frac{x}{2}\right) \\ \\
\frac{32}{17} &= 2\sin^2\left(\frac{x}{2}\right) \\ \\
\frac{16}{17} &= \sin^2\left(\frac{x}{2}\right) \\ \\
\pm\frac{4}{\sqrt{17}} &= \sin\left(\frac{x}{2}\right) \\ \\
\pm\frac{4\sqrt{17}}{17} &= \sin\left(\frac{x}{2}\right)
\end{align}
Now, we have to ask ourselves if the answer should be positive or negative, and why? If $x$ is in quadrant two, that means if we cut its angle in half, it will move to $x/2$ in quadrant 1. Try it yourself on a piece of paper. Or you can think about how the biggest angle in quadrant 2 is 180 degrees, so half of that is 90, so anything smaller than 180 must be cut in half to smaller than 90. 
In quadrant 1, sin values are positive so we know the answer has to be
\begin{align}\frac{4\sqrt{17}}{17} &= \sin\left(\frac{x}{2}\right)
\end{align}

Edit: don't forget the plus or minus when taking the square root.
</s>