Because they're not inverses, not exactly. $\sin x$ is not invertible on its entire domain because it's not one-to-one. There are infinitely many values of $x$ such that $\sin x=0$ , for example.

The standard function $\sin^{-1}x$ has a domain of $-1\le x\le1$ and outputs values between $-\dfrac\pi2$ and $\dfrac\pi2$ . This means that its inverse, $\sin x$ , is indeed its inverse as long as $-\dfrac\pi2\le x\le\dfrac\pi2$ .

$\dfrac{3\pi}4$ is larger than $\dfrac\pi2$ and thus does not fall in the "invertible part" of the domain of $\sin x$ . We have

$\sin\dfrac{3\pi}4=\dfrac1{\sqrt2}$

which is a value between -1 and 1, so that

$\sin^{-1}\left(\sin\dfrac{3\pi}4\right)=\sin^{-1}\left(\dfrac1{\sqrt2}\right)=\dfrac\pi4$

If we wanted to recover $\dfrac{3\pi}4$ , we'd have to redefine $\sin^{-1}x$ or define a new inverse function that works on a different branch of the domain of $\sin x$ .

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Fofino [41]

Answer:

Step-by-step explanation:

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