<s>[INST] Finding an angle which satisfies two equations I'd like to prove the following: Given any two real numbers $a$ and $b$, not both zero, there exists $c \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ such that $\sin c = \frac{a}{\sqrt{a^2 + b^2}}$ and $\cos c = \frac{b}{\sqrt{a^2 + b^2}}$. I'm trying to use as little geometry as possible, and appeal to the formal properties of sine and cosine.
So far what I have is this:

By hypothesis, $\sqrt{a^2 + b^2} \neq 0$. Thus, $\frac{a}{\sqrt{a^2 +b^2}} \in [-1,1]$,
  and it follows from the intermediate value theorem
  that there exists some $c \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ such
  that $\sin c = \frac{a}{\sqrt{a^2 + b^2}}$. But $\sin^2 c + \cos^2 c = 1$
  , which implies
$$  \cos^2 c = 1 - \frac{a^2}{a^2 + b^2} = \frac{b^2}{a^2 + b^2}. $$
It follows that $\cos c = \pm \frac{b}{\sqrt{a^2 + b^2}}$.

Of course, this is almost what I want, but is there a way for me to rule out the negative case? 
Thanks in advance for any help!
 [/INST] Your question is a trick question. 
It is impossible to prove such a claim because it is not true. If you really want to prove the truthness of such a claim I'd start with the claim that for any real numbers $a$ and $b$ there exists an angle $\theta$ such that $\cos{\theta} = \frac{a}{\sqrt{a^2 + b^2}}$, $\sin{\theta} = \frac{b}{\sqrt{a^2 + b^2}}$, and $ \theta \in [\frac{\pi}{2}, \frac{3\pi}{2}]$
A case where the claim you're trying to prove does not hold is when $a=-1$ and $b=0$
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