Evaluating $\sin^{-1}(\cos(40))$

Question

So as the title states I have to evaluate

$\sin^{-1}(\cos(40))$

In my textbook they answer it as following:

$\sin^{-1}(\cos(40))=90-\cos^{-1}(\cos(40))=50$

I'm however a little confounded over their answer.

As I recall this is the complementary angle identity but I don't really understand why it is used, considering is within the bound of [-1,1]

Would be extremely grateful if somebody could expand. If my question is unclear, I would be more than happy to further clearify.

Thank you in advance!

Answer 1

Does the following help?

$$\begin{align} \arcsin {\cos \theta}&=\omega \\ \cos \theta &=\sin \omega \\ \sin \left(\frac{\pi}{2}-\theta\right) &= \sin \omega \\ \end{align}$$

also there's this

$$\begin{align} \sin \theta &= \cos \left(\frac{\pi}{2}-\theta \right) \\ &\text{if} \ \ \theta =\arcsin \Omega \quad \ldots \quad \text{then} \\ \Omega&=\cos \left(\frac{\pi}{2}-\arcsin \Omega \right) \\ \arccos \Omega&=\frac{\pi}{2}-\arcsin \Omega \\ \color{red}{\arcsin \Omega} \ & =\color{red}{\frac{\pi}{2}-\arccos \Omega} \\ \frac{\pi}{2}&=\arcsin \Omega + \arccos \Omega \end{align}$$

Answer 2

Let $ \alpha =\sin^{-1}(\cos(40))$. That is $\sin(\alpha )=\cos(40)$. Using the formula $\sin( \alpha ) = \cos( \pi/2 -\alpha )$ results in $\alpha =50$ . Note that we are working in degree mode and using complementary angles.