I am trying to show that: $$(\sin(\theta) +i\cos(\theta))^n = \begin{cases} &\sin(n \theta) + i \cos(n\theta) & \quad \text{when $n = 4m+1$}\\ -&\sin(n \theta) - i \cos(n\theta) & \quad \text{when $n = 4m+3$} \end{cases} $$
The hint that I am given is to use the fact that $\sin$ and $\cos$ are the same if you translate one horizontally by $\pi/2$ radians, but I am not sure how this helps. Any hints, or solutions, would be appreciated.
Same hint in different phrasing: $$ \sin(\theta) + i\cos(\theta) = \cos(\pi/2-\theta) + i\sin(\pi/2-\theta) = e^{i(\pi/2-\theta)} $$
Alternatively, you can start by noting $$ \sin(\theta) + i\cos(\theta) = i\bigl(\cos(-\theta)+i\sin(-\theta)\bigr) $$