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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!

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    $\begingroup$ 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. $\endgroup$ – Jack D'Aurizio Jan 5 '18 at 2:10
  • $\begingroup$ How did you arrive at that conclusion? $\endgroup$ – oxodo Jan 5 '18 at 2:14
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    $\begingroup$ $\cos(40)=\sin\left(\frac{25\pi}{2}-40\right)$ and $\frac{25\pi}{2}-40$ is close enough to $0$ to be part of the range of the $\arcsin$ function. $\endgroup$ – Jack D'Aurizio Jan 5 '18 at 2:17
  • $\begingroup$ $\arcsin\cos(40^\circ)$ is the amplitude of the acute angle whose sine equals $\cos(40^\circ)=\sin(50^\circ)$, i.e. $50^\circ$. $\endgroup$ – Jack D'Aurizio Jan 5 '18 at 2:20
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    $\begingroup$ what exactly do you mean "it's within the bound of $[-1,1]$"? $\endgroup$ – MaximusFastidiousIrreverence Jan 5 '18 at 2:21
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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

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

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  • $\begingroup$ 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? $\endgroup$ – oxodo Jan 5 '18 at 2:53
  • $\begingroup$ 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 $\endgroup$ – MaximusFastidiousIrreverence Jan 5 '18 at 2:57
  • $\begingroup$ 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! $\endgroup$ – oxodo Jan 5 '18 at 3:06
  • $\begingroup$ They could have taken $$\cos \theta=\sin \omega \quad \to \quad \cos \theta=\cos \left( \frac{\pi}{2}-\omega \right)$$ instead? $\endgroup$ – MaximusFastidiousIrreverence Jan 5 '18 at 3:11
  • $\begingroup$ 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....? $\endgroup$ – MaximusFastidiousIrreverence Jan 5 '18 at 3:16
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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.

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  • $\begingroup$ 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 $\endgroup$ – oxodo Jan 5 '18 at 2:42
  • $\begingroup$ 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 )$? $\endgroup$ – Mohammad Riazi-Kermani Jan 5 '18 at 2:46
  • $\begingroup$ See if the edited version works better. $\endgroup$ – Mohammad Riazi-Kermani Jan 5 '18 at 2:57

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