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?
I don't see how the identity could be true. In the identity that you "vaugely recall", the only way for the right side to be identically zero is for the part in front of $sin$ to be zero. This requires that both $a$ and $b$ are zero, so $a^2+b^2$ would be zero.
The equation is homogeneous in $a$ and $b$, so that if $a,b$ fulfill it, so do $\lambda a,\lambda b$, where $\lambda$ is a free parameter. Then, there is no reason that $\lambda^2(a^2+b^2)=1$. On the opposite, the only way for $\lambda^2(a^2+b^2)$ to remain bounded is to be zero !