<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] It makes no sense for a mathematical claim to have a condition of the form "$-P-Q = 0$", so maybe the intended claim is:

Given reals $a,b,t$ such that $1 - a \sin(t) - b \cos(t) = 0$ and $a^2+b^2 \le 1$, then $a^2+b^2 = 1$.

And indeed your method proves this.
For fun, here is an alternative proof that uses weighted AM-GM:
Take reals $a,b,t$ such that $1 - a \sin(t) - b \cos(t) = 0$. Note that $a,b$ are not both zero.
Then $\dfrac1{a^2+b^2} = \dfrac{ a^2 \frac1a \sin(t) + b^2 \frac1b \cos(t) }{a^2+b^2} \le \sqrt{\dfrac{a^2 ( \frac1a \sin(t) )^2 + b^2 ( \frac1b \cos(t) )^2 }{a^2+b^2}} = \dfrac1{\sqrt{a^2+b^2}}$.
Thus $a^2+b^2 \ge 1$.
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