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Find the exact value of each expression.

(a) $ \cos^{-1} (-1) $(b) $ \sin^{-1} (0.5) $

(a) $\cos ^{-1}(-1)=\pi$ because $\cos \pi=-1$ and $\pi$ is in the interval $[0, \pi]$ (the range of $\cos ^{-1}$ ).(b) $\left.\sin ^{-1}(0.5)=\frac{\pi}{6} \text { because } \sin \frac{\pi}{6}=0.5 \text { and } \frac{\pi}{6} \text { is in the interval }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \text { (the range of } \sin ^{-1}\right)$.

01:25

Jeffrey P.

Calculus 1 / AB

Calculus 2 / BC

Calculus 3

Chapter 1

Functions and Models

Section 5

Inverse Functions and Logarithms

Functions

Integration Techniques

Partial Derivatives

Functions of Several Variables

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When you look at this expression inverse coastline of negative one, here's what you want to think. It's asking you to find the angle. Who's co sign is negative one. And remember that for the inverse coastline function were limited to a range of outputs between zero and pie. So if you think about your unit circle over here when you've gone Pie radiance, you reach the point Negative 10 so the coastline is negative one. So the answer to that is pie. Similarly, for the ember sign of 1/2 when you see that phrase you want to think to yourself, you're being asked to find the angle. Every time you're doing an inverse trig function, you're finding the angle. Who's sign is 1/2? I'm going to think of it as 1/2 rather than 0.5. And remember, when dealing with Enver sign, your outputs have to fall in the range between negative pi over to two pi over two. So for positive 1/2 we think of this In Quadrant one, the sign is 1/2 so the opposite would be one, and the high pot news would be to, and hopefully you recognize that as a 30 60 90 triangle with a 30 degree reference angle. So the answer is 30 degrees, although we should give her answer and radiance. So instead of thinking of it as 30 degrees, you think of it as pi over six.

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