Finding local inverse of $y = \sin x$ at the point $x = 5\pi/6$ and at the point $x = 7\pi/6$

Question

Find a formula for the local inverse to the function $f: \mathbb{R} \to \mathbb{R}$ defined by

$$y = \sin(x)$$

at the point $x=5\pi/6$ and at the point $x=-7\pi/6$.

Can anybody help me with this exercise?

Answer 1

The function $f: \mathbb{R} \to [-1, 1]$ defined by $f(x) = \sin x$ is periodic. It does not have an inverse since for each $y$ in its range, there are infinitely many values of $x$ such that $f(x) = y$. For instance, if $y = 0$, then $x = n\pi, n \in \mathbb{Z}$.

Consequently, to obtain an inverse, we must restrict the domain of the sine function so that there is one value of $x$ for each value of $y \in [-1, 1]$. We can do this for any interval in which the sine function assumes all values in the range exactly once. By convention, the standard restriction is to the interval $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, which yields the function $$g: \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \to [-1, 1]$$ defined by $$g(x) = \sin x$$

The inverse of $g$ is the arcsine function $$h: [-1, 1] \to \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$ defined by $$\arcsin x = y \iff x = \sin y, -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$$

However, $5\pi/6$ does not lie in the range of $h(x) = \arcsin x$. To ensure that it does lie in the range, we must find a different restriction on the domain of $f$. Observe that $f(x) = \sin x$ assumes every value in $[-1, 1]$ over the interval $\left[\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right]$, that each value in the range occurs exactly once over the interval $\left[\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right]$, and that $\dfrac{5\pi}{6} \in \left[\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right]$. Thus, if we define the function $$k: \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \to [-1, 1]$$ by $$k(x) = \sin x$$ then the function $k$ will have inverse $$m: [-1, 1] \to \left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$$ defined by $$m(x) = y \iff \sin y = x, \frac{\pi}{2} \leq x \leq \frac{3\pi}{2}$$ In particular, $$m\left(\frac{1}{2}\right) = \frac{5\pi}{6}$$

The function $m$ is your desired local inverse for $f$ at $x = \dfrac{5\pi}{6}$. I will leave it to you to determine the local inverse for $f$ at $x = -\dfrac{7\pi}{6}$.