<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] 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.)
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