A question about composition of tigonometric functions

**yotamoo** · November 8th 2011, 01:45 AM

A little something I'm trying to understand:

sin(arcsin(x)) is always x, but
arcsin(sin(x)) is not always x

So my question is simple - why? Since each cancels the other, it would make sense that arcsin(sin(x)) would always result in x.
I'd appreciate any explanation. Thanks!

**Prove It** · November 8th 2011, 01:57 AM

Originally Posted by yotamoo

A little something I'm trying to understand:
So my question is simple - why? Since each cancels the other, it would make sense that arcsin(sin(x)) would always result in x.
I'd appreciate any explanation. Thanks!

Because adding any integer multiple of $\displaystyle 2\pi$ also satisfies the solution to the second...

**Soroban** · November 8th 2011, 08:13 PM

Hello, yotamoo!

A little something I'm trying to understand:

. . $\text{(a) }\sin(\arcsin x)$ is always $x.$

. . $\text{(b) }\arcsin(\sin x)$ is not always $x.$

So my question is: why?

I'll illustrate with a specific example.

(a) We have: . $\sin(\arcsin x)$

Suppose $x = \tfrac{1}{2}$

Then: . $\arcsin\left(\tfrac{1}{2}\right) \:=\:\begin{Bmatrix}\frac{\pi}{6} + 2\pi n \\ \\[-4mm] \frac{5\pi}{6} + 2\pi n \end{Bmatrix}$

Hence: . $\sin\begin{pmatrix} \frac{\pi}{6} + 2\pi n \\ \frac{5\pi}{6} + 2\pi n \end{pmatrix} \:=\:\tfrac{1}{2}$

We get back our original value of $x.$

(b) We have: . $\arcsin(\sin x)$

Suppose $x = \tfrac{5\pi}{6}$

Then: . $\sin\tfrac{5\pi}{6} \,=\,\tfrac{1}{2}$

Hence: . $\arcsin\left(\tfrac{1}{2}\right) \;=\;\begin{Bmatrix}\frac{\pi}{6} + 2\pi n \\ \frac{5\pi}{6} + 2\pi n \end{Bmatrix}$

There is an infinite number of possible values.
We do not get back our original value of $x.$

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