For $0\leq a,b \leq \pi$ if $\cos a = -\cos b$ then $\sin a = \sin b$ Problem:
Is it it true that for $0\leq a,b \leq \pi$ 
if $\cos a = -\cos b$ then $\sin a = \sin b$.
Thoughts:
When I draw a picture it makes sense but I am unsure what identity to use to show this is true. Any hints appreciated.
What about this:
First squaring both sides yields
$\cos^2(a) = \cos^2(b)$, then $1-\sin^2(a) = 1 - \sin^2(b)$ implies $\sin(a) = \sin(b)$
 A: $\cos(a) = -\cos(b) \Rightarrow \cos(a)=\cos(\pi-b)\Rightarrow a=2k\pi\pm (\pi-b)$
Since $0\leq a, b\leq \pi$, we can understand that the aforementioned inequality works only for $k=0$ and $+(\pi - b)$:
$\Rightarrow a = \pi - b \Rightarrow \sin(a) = \sin(b)$ 

I'm using the fact that when
$\sin(x) = \sin(a) \Rightarrow x = 2k\pi +a$ or $2k\pi + \pi - a$
$\cos(x) = \cos(a) \Rightarrow x = 2k\pi \pm a$
$\tan(x) = \tan(a) \Rightarrow x = k\pi + a$
A: Your idea is good, but what you have doesn't imply that $\sin a = \sin b$, but $\sin a = \pm \sin b$. Luckily, we can easily remedy this:
$$\cos a=-\cos b\implies \sin^2a = 1-\cos^2a=1-\cos^2b = \sin^2b\implies |\sin a\,| = |\sin b\,|,$$
and since $\sin x\geq 0$ for $x\in[0,\pi]$, we have $$\sin a =|\sin a\,| = |\sin b\,| = \sin b.$$
A: Note:
$$ \cos b = -\cos(\pi - b) = -\cos(\pi+b) $$
Thus
$$ \begin{align}
\cos a &= -\cos b, \quad (1)\\
\cos a &= \begin{cases}
 \cos(\pi +b) \\
 \cos(\pi -b)
\end{cases}\\
\implies a &= \begin{cases}
 \pi +b + 2n\pi \\
 \pi -b + 2n\pi
\end{cases}
\end{align}
$$
where $n= 0, \pm 1, \pm 2, \ldots.$
Imposing the condition $0 \le a,\, b \le \pi$ reduces the set of solutions above to just a single solution:
$$ a = \pi -b$$
Hence, it follows that
$$\sin a = \sin b.$$
Note: Don't square both sides of $(1)$ because doing so will introduce spurious solutions and, hence, requires an additional step to verify all solutions at the end.
