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Find the exact value of each expression.

(a) $ \csc^{-1} \sqrt{2} $

(b) $ \arcsin 1 $

(a) $\left.\csc ^{-1} \sqrt{2}=\frac{\pi}{4} \text { because } \csc \frac{\pi}{4}=\sqrt{2} \text { and } \frac{\pi}{4} \text { is in }\left(0, \frac{\pi}{2}\right] \cup\left(\pi, \frac{3 \pi}{2}\right] \text { (the range of } \csc ^{-2}\right)$.

(b) arcsin $1=\frac{\pi}{2}$ because $\sin \frac{\pi}{2}=1$ and $\frac{\pi}{2}$ is in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (the range of arcsin).

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Johns Hopkins University

Missouri State University

Harvey Mudd College

Idaho State University

when you see an expression like this inverse cosi can't of square root. Two. Here's what you want to go through your mind that's asking you to find the angle whenever doing inverse Trig. You're finding the angle whose coast he can't is square root, too, and you could think of square root to a square to over one. Now remember, the angles have to fall, either in Quadrant four between negative pi over two and zero or in quadrant one between zero and pi over two. And because we have a positive value. Positive Square root two were in quadrant one. So let's draw reference Triangle in quadrant one. And what we know is the coast he can't is square root to over one coast he can't is high pot noose over opposite high partners over opposite, so we can put square root to on the high pot news and one on the opposite. Hopefully, that's enough to show us what kind of special right triangle we have. The other leg would be one, and we should recognize that as a 45 45 90 triangle, so the angle is 45 degrees. The angle that we're looking for, but we're not looking forward in degrees. We're looking forward in radiance. So that would be pi over four radiance. Similarly, we have arc sine of one and Arc sign means inver sign is just a different way of saying it. So when you see that what you want to think is is asking you to find the angle again. You're finding an angle every time you're doing inverse Trig, find the angle. Whose sign is one? Now remember that your outputs for arc sine have to fall between negative pi over two and pi over two. That would be quadrants one and four. Where is the sign? Equal to one. We're gonna have to go to our unit circle for this one. And remember that up here at 90 degrees, the point has coordinates 01 So the sign value is one. So that's 90 degrees. Although we don't want to answer in degrees, we want to answering radiance. So convert that to radiance and you get pi over two