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

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.

• If you are using degrees, please make it clear by using $40^\circ, 50^\circ$ and $90^\circ$. I would say that $$\arcsin\cos(40)=-40 +\frac{25\pi}{2},$$ otherwise. Jan 5, 2018 at 2:10
• How did you arrive at that conclusion? Jan 5, 2018 at 2:14
• $\cos(40)=\sin\left(\frac{25\pi}{2}-40\right)$ and $\frac{25\pi}{2}-40$ is close enough to $0$ to be part of the range of the $\arcsin$ function. Jan 5, 2018 at 2:17
• $\arcsin\cos(40^\circ)$ is the amplitude of the acute angle whose sine equals $\cos(40^\circ)=\sin(50^\circ)$, i.e. $50^\circ$. Jan 5, 2018 at 2:20
• what exactly do you mean "it's within the bound of $[-1,1]$"? Jan 5, 2018 at 2:21

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}

• Right. So could I reason as follows: $cos^\circ{40}=sin(\frac{pi}{2}-40^\circ)$ and $sin^{-1}(sin(\frac{pi}{2}-40^\circ)=(\frac{pi}{2}-40^\circ)$ Because of the cancellation identities? Jan 5, 2018 at 2:53
• yes. It's the same as in the above taking the arcsine of both sides an arriving at $$\frac{\pi}{2}-\theta=\omega$$ such that theta is $40$ degrees and omega is the answer you seek Jan 5, 2018 at 2:57
• Alright! thank you. I'm still unsure how the author arrived at $90-\cos^{-1}(\cos(40))$ specifically $\cos^{-1}(\cos(40))$ all I know is that he applied thecomplementary angle identity to get: $\sin^{-1}(\cos(40))=90-\cos^{-1}(\cos(40))=50$ Anyhow, thanks! Jan 5, 2018 at 3:06
• They could have taken $$\cos \theta=\sin \omega \quad \to \quad \cos \theta=\cos \left( \frac{\pi}{2}-\omega \right)$$ instead? Jan 5, 2018 at 3:11
• This identity: $$\arcsin x=\frac{\pi}{2} - \arccos x$$ may more directly apply here. Set $\quad x=\cos{\theta} \quad$. The book likely just resorted to that....? Jan 5, 2018 at 3:16

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.

• Right, that makes sense. But how would you arrive to the author's answer if you would've to write it down? Thank you btw Jan 5, 2018 at 2:42
• It depends on what is known and what is not known at this course. Can we use formulas like$sin(\pi /2 - \alpha )= cos(\alpha )$? Jan 5, 2018 at 2:46
• See if the edited version works better. Jan 5, 2018 at 2:57