Given $y=\arccos(x)$ find $\arcsin(x)$ in terms of y
By Abigail Rogers •
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
$\endgroup$3 Answers
$\begingroup$HINT: $$x=\cos y=\sin\left(\frac{\pi }{2}-y\right)$$
$\endgroup$ 0 $\begingroup$$$y = \arccos x \iff \cos y = x \iff x = \sin\left(\frac \pi 2 - y\right)\iff \arcsin x = \left(\frac \pi 2 - y\right)$$
$\endgroup$ 1 $\begingroup$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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