$p_0^2 + q_0^2 = 1$ and $-\sin(s) = -p_0\sin(s) + q_0\cos(s)$ $\rightarrow $ $p_0 = -\cos2s, q_0 = -\sin(2s)$

Question

How does solving $p_0^2 + q_0^2 = 1$ and $-\sin(s) = -p_0\sin(s) + q_0\cos(s)$ give $p_0 = -\cos2s $ and $q_0 = -\sin(2s)$?

I can clearly see one solution is $p_0=1,q_0=0$ but can't seem to understand the other.

Answer 1

HINT : Let $p_0=\cos x $ and $q_0=\sin x$. Then the second equation becomes $$-\sin s = -\cos x\sin s + \sin x\cos s.$$ That is $$-\sin s = \sin(x-s).$$ Now solve for $x$.

Answer 2

Since $p_0^2 + q_0^2 = 1$, there is $\theta$ such that $p_0 = \cos(\theta)$ and $q_0 = \sin(\theta)$. Thus your equation becomes $$ \sin(-s) = -\sin(s) = - \cos(\theta) \sin(s) + \sin(\theta) \cos(s) = \sin(\theta - s)$$ Thus either $\theta = 2 n \pi$ or $\theta = (2n+1)\pi + 2 s$.