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Thread: A question about composition of tigonometric functions

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    A question about composition of tigonometric functions

    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!
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    Re: A question about composition of tigonometric functions

    Quote Originally Posted by yotamoo View Post
    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 $\displaystyle \displaystyle 2\pi$ also satisfies the solution to the second...
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    Re: A question about composition of tigonometric functions

    Hello, yotamoo!

    A little something I'm trying to understand:

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

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

    So my question is: why?

    I'll illustrate with a specific example.


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

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

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

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

    We get back our original value of $\displaystyle x.$



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

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

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

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

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

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