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We are given trigonometry expression.

[tex]\sin^{-1}(\sin(\frac{3\pi}{4}))\neq \frac{3\pi}{4}[/tex]

when inverse of [tex]y=\sin x[/tex] is [tex]y=\sin^{-1}x[/tex]

Inverse function gives principle value of angle.

Here we have sin inverse function whose principle value is [tex]-\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2}[/tex]

The value of y must be lie between principle value. This changes because of exact value of sine. If we find [tex]\sin(3\pi/4)=\frac{1}{\sqrt{2}}[/tex]

As we know [tex]\sin(\frac{\pi}{4})=\frac{1}{\sqrt{2}}[/tex]

Thus, The principle value are not same.

Hence, [tex]\sin^{-1}(\sin(\frac{3\pi}{4}))\neq \frac{3\pi}{4}[/tex]

Inverse functions when applied sequentially, cancel out the effect of transformation. The result obtained is because of cancellation of sin function.

What is inverse function of a function?

Suppose that there is a function [tex]y = f(x)[/tex], such that its both one-to-one and onto function. Then, its inverse function is defined as [tex]y = f^{-1}(x)[/tex] such that:

[tex]f^{-1}(f(x)) = x[/tex] (for all x from domain of f)

This is the reason, why we get:

[tex]\sin^{-1}(\sin(\dfrac{3\pi}{4}))=\dfrac{3\pi}{4}[/tex]

It is because it was given that [tex]y = sin(x)[/tex] and [tex]y=sin^{-1}x[/tex] are inverses (of each other).

Actually, this inverse is defined in a way that we're only taking principal values(restricted to some domain) (and not general values) of the angle.

Thus,

The result obtained is because of cancellation of sin function due to its inverse function.

Learn more about inverse of sin here:
https://brainly.com/question/6826226

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