Simplify $\arcsin (\cos (2x)), 0 \leq x \leq (\pi/2)$
My answer:
I used the identity $\sin ((\pi/2) - x) = \cos x$ and I got $(\pi/2) - 2x$. Am I correct?
Simplify $\arcsin (\cos (2x)), 0 \leq x \leq (\pi/2)$
My answer:
I used the identity $\sin ((\pi/2) - x) = \cos x$ and I got $(\pi/2) - 2x$. Am I correct?
Copyright © 2021 JogjaFile Inc.
$$\arcsin(\cos2x)=\dfrac\pi2-\arccos(\cos2x)$$
As $0\le2x\le\pi$ using Principal values
$$\arccos(\cos2x)=2x$$
If $\pi<2x\le2\pi,\arccos(\cos2x)=2\pi-2x$
If $-\pi<2x<0,\arccos(\cos2x)=-2x$