This is part of a larger problem that I'm doing but I've gotten stuck on this part. I set up a right triangle to help me (with 1 as the hypotenuse, x as the side opposite the angle, and square root 1-x^2 as the adjacent side), but I was wondering how that triangle would change given that the question isn't asking for sin(arcsinx), but rather sin(arcsin1/x). Any help about how to move forward with this would be appreciated! Thank you!
How to evaluate sin(arcsin1/x) exactly?
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Dr.Peterson
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It doesn't matter what the argument is; sin(arcsin(u)) always equals u, whatever u is -- as long as u is in the domain of the arcsin. So the latter is the only thing you may have to check.
And for this one, you don't need to draw that triangle, as you might if the functions weren't a matched pair. You can read the expression as "the sin of the angle between -pi/2 and pi/2 whose sine is 1/x." That's not much different from "Who is buried in Grant's tomb?"
Thank you so much, that property totally slipped my mind. I also have to solve cos(sin1/x) for the problem as well. How should I go about doing that part of the problem?
Dr.Peterson
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I hope you mean cos(sin^{-1}(1/x)) ! That means, "What is the cosine of an angle (between -pi/2 and pi/2) whose sine is 1/x?"
That's where you can use the triangle (or identities, if you prefer). I would put the 1 on the opposite side and x on the hypotenuse.
Yes, I mean cos(sin^-1(1/x)) (sorry for the typo), and thank you so much for the help!