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. 
 A: Guide:
We do not plug in a value of $x$, we solve for the trigonometry value without having the need to find its value explicitly.
If $x$ is in the second quadrant, then $\frac{x}2$ is in the first quadrant. 
$$\cos(x) = -\frac{15}{17}$$
is $$2\cos^2\left( \frac{x}2\right)-1=-\frac{15}{17}$$
Solve for $\cos\left( \frac{x}2\right)$.
For the second question, given $\sin(x)$ and knowing which quadrant it is in, you should be able to find $\cos(x)$ using Pythagoras Theorem, and use double angle formula to recover $\sin(2x), \cos(2x)$. Then use the definition of tangent.
A: It is very helpful to know the following identities.
1)  $\sin ^2 x + \ cos ^2 x =1 $ 
2) $\sin ^2 x = (1/2)(1-\cos 2x)$
3) $\cos ^2 x = (1/2 )(1+ \cos 2x)$
4) $\sin 2x = 2\sin x \cos x$
5) $\cos 2x = \cos ^2 x - \sin ^2x $
A: *

*You have formulas, namely:


$sin(2x) = 2*cos(x)*sin(x)$
$cos(2x) = cos^2(x) - sin^2(x)$
$sin^2(x) + cos^2(x) = 1$
There are plenty more, but generally speaking everything can be derived from each other.


*You have the unit circle, which helps you determine the sign of the function. You need it, because when you try to derive a, for example, sin from the trig identity, you end up with expression: $ |sin(x)| = \sqrt{1 - cos^2(x)}$ and at this point the quadrant information helps you recover the sign of sin(x)


For the first question you are given that $cos(x) = -15/17$ and $x$ is in the 2nd quadrant. This tells you, that $x \in [\pi/2;\pi]$. Respectively, $x/2 \in [\pi/4;\pi/2] = $ first quadrant, which means, that cos(x/2) and sin(x/2) are positive.
Now you just use the formula, for example, the second one I listed, to find the cos(x/2)
$cos(x) = cos^2(x/2) - sin^2(x/2) = 2cos^2(x/2) - 1 => |cos(x/2)| = cos(x/2)$(verified sign already to be positive)$ = \sqrt{\frac{1 + cos(x)}{2}}$
I guess you can do the math.
A: 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.
A: For your first question, use the half-angle formulas.
$$\sin\frac{x}{2} = \pm\sqrt{\frac{1-\cos x}{2}}$$
$$\cos\frac{x}{2} = \pm\sqrt{\frac{1+\cos x}{2}}$$
$\sin\frac{x}{2}$ and $\cos\frac{x}{2}$ can take positive or negative values depending on which Quadrant $\frac{x}{2}$ is in.
Since $x$ is in Quadrant $II$, then $\frac{\pi}{2} \leq x \leq \pi$. Divide everything by $2$ to get $\frac{\pi}{4} \leq \frac{x}{2} \leq \frac{\pi}{2}$. 
Notice that the angle is is between $\frac{\pi}{2}$ and $\frac{\pi}{4}$, so the half-angle must be in Quadrant $I$. Therefore, $\sin \frac{\pi}{2}$ and $\cos \frac{\pi}{2}$ must take positive values. Use $\cos x = -\frac{15}{17}$.
$$\sin\frac{x}{2} = +\sqrt{\frac{1-\big(-\frac{15}{17}\big)}{2}}$$
$$\implies \sin\frac{x}{2} = +\sqrt{\frac{\frac{32}{17}}{2}}$$
$$\implies \sin\frac{x}{2} = +\sqrt{\frac{32}{34}}$$
$$\implies \sin\frac{x}{2} = +\sqrt{\frac{16}{17}}$$
$$\implies \boxed{\sin\frac{x}{2} = +\frac{4}{\sqrt{17}} = +\frac{4\sqrt{17}}{17}}$$
Do the same thing for finding its cosine value.
The second question is very similar. Use the double-angle formulas. (For cosine, any of the given equations can be used.)
$$\sin 2x = 2\sin x\cos x$$
$$\cos 2x = \cos^2 x-\sin^2 x = 2\cos^2 x-1 = \color{blue}{1-2\sin^2 x}$$
$$\tan 2x = \color{blue}{\frac{\sin 2x}{\cos 2x}} = \frac{2\tan x}{1-\tan^2 x}$$
Use $\sin x = -\frac{7}{25}$. Considering you’re given $\sin x$, the blue formulas are the best to use. (Otherwise you’d have to compute $\cos x$ and $\tan x$ first.)
