Why is $\int \cos(n\arccos(\sin(x)) = \int \cos(nx)$ on $[-1, 1]$?

Question

I would like to evaluate

$$\int \cos(n\arccos(\sin(x)) dx$$

with $x \in [-1, 1],$ and I used a calculator do to so. At the very beginning, the calculator simplifies

$$\int \cos(n\arccos(\sin(x)) dx = \int \cos(nx) dx.$$

This looks like $\arccos(\sin(x)) = x$, but I don't see where this comes from.

Note that if $y=\arccos(\sin(x))$, then $\cos(y)=\sin(x)$. – Mark Viola Jul 01 '16 at 16:47 — Mark Viola, Jul 01 '16 at 16:47

b00n heT · Accepted Answer · 2016-07-01T18:21:55.847

1

Not a full answer, but couldn't fit it in a comment.

As I wrote above, recalling that $\sin(x)=\cos(\pi/2-x)$ we obtain that \begin{align*}\cos(n\cdot \arccos(\sin(x)))& =\cos(n\cdot\arccos(\cos(\pi/2-x)))=\cos(n\pi/2-nx )\\& =\cos(n\pi/2)\cos(nx)+\sin(n\pi/2)\sin(nx)\end{align*} which is equal to $n\cos(x)$ iff $n=0\mod 4$.

Are we sure that the substitution is correct?

edited Jul 01 '16 at 18:21

answered Jul 01 '16 at 16:55

b00n heT

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You have a typo: $\cos(n\cdot\arccos(\cos(\pi/2-x)))=\cos(n\pi/2-n x )$. – Emilio Novati Jul 01 '16 at 17:25
You are right! Thank you – b00n heT Jul 01 '16 at 18:21

Why is $\int \cos(n\arccos(\sin(x)) = \int \cos(nx)$ on $[-1, 1]$?

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