<s>[INST] Suppose $-a\sin(s) - b\cos(s) = 0$, then $a^2 + b^2 = 1$? 
Suppose $-a\sin(s) - b\cos(s)  = 0$ and given that $a^2 + b^2 \le 1$, then $a^2 + b^2 = 1$?

I am having trouble getting the above identity. I vaguely recall that
$$a\sin(s) + b\cos(s)=\sqrt{a^2+b^2}\sin(s+\theta)$$
where $$\tan(\theta) = \frac{b}{a}$$
But I still don't see why it's obvious?
 [/INST] Let us see the identity you vaguely recall:
$$a\sin(s) + b\cos(s)=\sqrt{a^2+b^2}\sin(s+\theta)$$
with $\tan\theta=\frac ba$. To get it, we factor out the square root, then set $\sin\theta=\frac{b}{\sqrt{a^2+b^2}}$ and $\cos\theta=\frac{a}{\sqrt{a^2+b^2}}$, which is licit since the squares of those numbers sum to 1. Now we have:
$$a\sin s+b\cos s=\sqrt{a^2+b^2}(\cos\theta\sin s+\sin\theta\cos s),$$
and $\sin(s+\theta)$ expands to just that using addition formula.
So the identity at question top now becomes:
$$\sqrt{a^2+b^2}\sin(s+\theta)=0.$$
I have, of course, changed signs, which doesn't alter the equation. This can never be true for all $s$ unless $a=b=0$, in which case obviously $a^2+b^2=0\neq1$.
Another answer suggests the problem might have a $1$ at the start of the equations, and that the identity works only for some $t$. Then we would have, for some $t$, that:
$$1-\sqrt{a^2+b^2}\sin(t+\theta)=0\iff\sin(t+\theta)=\frac{1}{\sqrt{a^2+b^2}}.$$
$a^2+b^2\leq1$, so the RHS is at least 1, but the LHS is at most 1 in absolute value, so for the equation to hold you need $a^2+b^2=1$.
</s>