Simplify $\arcsin (\cos (2x)), 0 \leq x \leq (\pi/2)$.

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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?

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$$\arcsin(\cos2x)=\dfrac\pi2-\arccos(\cos2x)$$

As $0\le2x\le\pi$ using Principal values

$$\arccos(\cos2x)=2x$$

If $\pi<2x\le2\pi,\arccos(\cos2x)=2\pi-2x$

If $-\pi<2x<0,\arccos(\cos2x)=-2x$