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Given that $y = \arccos x$, $ - 1 \le x \le 1\,and\,0 \le y \le \pi $, express $\arcsin x$ in terms of y.

The best I know how to do this is is:

$$\eqalign{ & \cos y = x \cr & {\cos ^2}y + {\sin ^2}y = 1 \cr & {\sin ^2}y = 1 - {\cos ^2}y \cr & \sin y = \sqrt {1 - {x^2}} \cr & \arcsin \left( {\sqrt {1 - {x^2}} } \right) = y \cr} $$

However this isn't what is asked for, How do I go about getting things in terms of y?

Thanks

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3 Answers 3

up vote 3 down vote accepted

HINT: $$x=\cos y=\sin\left(\frac{\pi }{2}-y\right)$$

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Thanks, ive got it now –  seeker Jun 12 '13 at 16:50

$$y = \arccos x \iff \cos y = x \iff x = \sin\left(\frac \pi 2 - y\right)\iff \arcsin x = \left(\frac \pi 2 - y\right)$$

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@Amzoti hehehehe... yes indeed! –  amWhy Jun 13 '13 at 0:29

Hint: Draw a right triangle with angle $y$. Let the hypotenuse be $1$ and the adjacent side $x$. Now what is the opposite side? What is $\sin y$?

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