Show that $\cos^{-1}\cos x=\sin^{-1}\sin \left(x-\dfrac{\pi}{2}\right)+\dfrac{\pi}{2}$

There are two ways to see this result.

Graphical 'proof'

One is sketch the graphs and as they are both zigzags it is easy to see the horizontal and vertical translations through $\dfrac{\pi}{2}$, see diagrams

Algebraic proof

Expand,

$\begin{array}{rl}\sin^{-1}\sin \left(x-\dfrac{\pi}{2}\right) &= \sin^{-1}(\sin x \cos\dfrac{\pi}{2}-\cos x\sin\dfrac{\pi}{2})\\&=\sin^{-1}(-\cos x)\\&=-\sin^{-1}\cos x\end{array}$

But we know that $$\sin^{-1}p+\cos^{-1}p=\dfrac{\pi}{2}$$ for all $p$.

$\therefore \cos^{-1}\cos x+\sin^{-1}\cos x=\dfrac{\pi}{2}$

$\therefore \sin^{-1}\sin \left(x-\dfrac{\pi}{2}\right) = \cos^{-1}\cos x -\dfrac{\pi}{2}$

$\therefore \cos^{-1}\cos x = \sin^{-1}\sin \left(x-\dfrac{\pi}{2}\right) +\dfrac{\pi}{2}$