Given the continuous curve $y=\sin \left(x\right)$ restricted to domain $[-\frac{\pi }{2},\frac{\pi }{2}]$ and codomain $[-1,1]$, evaluate whether the restricted function has an inverse on the codomain, and if so, what is an explicit form for that inverse.

• A. No, sinusoidal functions never have inverses on any interval because they oscillate; ${\sin }^{-1}$ is not a true inverse
• B. Yes, the inverse exists and is $f^{-1}\left(y\right)=\sin \left(y\right)$ because sine is its own inverse when restricted to any symmetric interval
• C. Yes, the restriction is one-to-one and has inverse $f^{-1}\left(y\right)={\sin }^{-1}\left(y\right)$ with domain $[-1,1]$ and range $[-\frac{\pi }{2},\frac{\pi }{2}]$
• D. No, because the codomain $[-1,1]$ is too large for an inverse; you must restrict the codomain to $\left\{0\right\}$ to have an inverse

Answer: (C)