Question

PLEASE HELP ASAP NOW WILL MARK BRAINIEST LOTS OF POINTS!!!!

Answer 1

22. On its own, $\sin x$ is not invertible because it's not one-to-one because it's periodic. For instance, we can always find more than one value of $x$ for which $\sin x=0$; this happens when $x=0,\pm\pi,\pm2\pi,\ldots$.

But we can restrict the domain so that it can become invertible. If we only allow values of $x$ within $-\dfrac\pi2\le x\le\dfrac\pi2$, for example, then each $x$ will only be associated with a single value of $\sin x$. This is how the standard inverse sine is defined. With $-\dfrac\pi2\le x\le\dfrac\pi2$ (restricted domain of sine), we guarantee that $-1\le\sin x\le1$ (range). So the domain of the inverse is $-1\le x\le 1$, and the range of the inverse is $-\dfrac\pi2\le\sin^{-1}x\le\dfrac\pi2$.

$\dfrac{3\pi}4$ is not in this restricted domain. But $\sin\dfrac{3\pi}4$ still exists as long as we take the standard domain (the entire real line), and $\sin\dfrac{3\pi}4=\dfrac1{\sqrt2}$. But then $\sin^{-1}\left(\sin\dfrac{3\pi}4\right)=\dfrac\pi4$ because this is the only value of $x$ for which $\sin^{-1}x=\dfrac1{\sqrt2}$.

In short: $\sin$ and $\sin^{-1}$ are NOT inverses of one another, but rather one is an imperfect inverse of the other.

25. Not much to say about this:

$g(x)=\cos x\implies g\left(\dfrac\pi2\right)=\cos\dfrac\pi2=0$

$f(x)=\sin x\implies f\left(g\left(\dfrac\pi2\right)\right)=\sin\left(\cos\dfrac\pi2\right)=\sin0=0$