Why does $\cos\theta=\sin(\frac{\pi}{2}-\theta)$ even when $\theta>90^\circ$?

Question

I get why this relationship holds true in $90$ degree triangles, but I don't get why this relationship holds true when $\theta>90^\circ$? I get that when $\theta>90^\circ$, you find the acute angle that describes $\theta$, if you see what I mean. But there is no way of showing that in this function! So why would the fact that $\cos\theta=\sin(\frac{\pi}{2}-\theta)$ still hold true here?

Answer 1

Draw a centre-$O$ circle of radius $1$ and add in the radius with an end at $(1,\,0)$, and rotate that radius through an angle $\theta$ anticlockwise so its end becomes $(\cos\theta,\,\sin\theta)$. You could have got the same result by rotating the radius at $(0,\,1)$ clockwise through $\frac{\pi}{2}-\theta$, so $(\cos\theta,\,\sin\theta)=(\sin(\frac{\pi}{2}-\theta),\,\cos(\frac{\pi}{2}-\theta))$.

Answer 2

Even a student that has only learned the ‘geometric definition’ of trigonometric functions can derive $$\cos x=\sin(\frac\pi{2}-x)$$ for $0<x<\frac\pi2$.

Then one would learn the Maclaurin series of $\sin$ and $\cos$, and then one can show that they have an infinite radius of convergence.

Since $(0,\frac{\pi}2)\subset\mathbb C$ and $\cos x=\sin(\frac\pi2-x)$ in this interval, by identity theorem this immediately implies the two functions($\cos x$ and $\sin(\frac\pi2-x)$) are equal for any $x\in \mathbb C$.