Exercise 2.2 · Q10: $\sin^{-1}\left(\sin\frac{2\pi}{3}\right)$

Basics you need

Solution

1. Why the answer is not $\frac{2\pi}{3}$

   Whatever $\sin^{-1}$ returns lies in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$⁠, that is, between $-90^\circ$ and $90^\circ$⁠. But

   $$\frac{2\pi}{3}=120^\circ>90^\circ=\frac{\pi}{2},$$

   so $\frac{2\pi}{3}\notin\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$⁠. Hence $\sin^{-1}\left(\sin\frac{2\pi}{3}\right)\ne\frac{2\pi}{3}$⁠.

   Here is the reason in full. Sine is not one-one: $\frac{\pi}{3}$ and $\frac{2\pi}{3}$ (and infinitely many other angles) all have sine $\frac{\sqrt3}{2}$⁠. A function cannot send $\frac{\sqrt3}{2}$ back to all of them, so $\sin^{-1}$ is defined to return just one: the one in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$⁠. So $\sin^{-1}$ undoes $\sin$ only for inputs that started in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$⁠. An angle like $\frac{2\pi}{3}$⁠, from outside, comes back as its partner inside the interval.
   $\sin^{-1}(\sin x)=x$ is stated in Section 2.3 only "for $x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$". Basics: what an inverse needs

2. Rewrite $\sin\frac{2\pi}{3}$ using an angle in the principal range

   $$\begin{aligned} \sin\frac{2\pi}{3} &= \sin\left(\pi-\frac{\pi}{3}\right) \\ &= \sin\frac{\pi}{3}=\frac{\sqrt3}{2}. \end{aligned}$$

   And $\frac{\pi}{3}=60^\circ$ does lie in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$⁠.

   We want an angle with the same sine as $\frac{2\pi}{3}$ but inside the principal range. The identity $\sin(\pi-u)=\sin u$ produces it: reflect $\frac{2\pi}{3}$ across the vertical axis of the unit circle to get $\frac{\pi}{3}$⁠. The two points are at the same height, so they have the same sine.
   $\sin(\pi-u)=\sin u$ (Class XI, Chapter 3). Basics: special angles and reference angles

3. Take $\sin^{-1}$

   $$\begin{aligned} \sin^{-1}\left(\sin\frac{2\pi}{3}\right) &= \sin^{-1}\left(\sin\frac{\pi}{3}\right) \\ &= \frac{\pi}{3}, \end{aligned}$$

   where the last equality uses $\sin^{-1}(\sin x)=x$⁠, valid because $\frac{\pi}{3}\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$⁠. Equivalently, $\sin^{-1}\frac{\sqrt3}{2}=\frac{\pi}{3}$⁠, since $\frac{\pi}{3}$ is the only angle in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ with sine $\frac{\sqrt3}{2}$⁠.

   This is the same method the book uses in Example 6 for $\sin^{-1}\left(\sin\frac{3\pi}{5}\right)=\frac{2\pi}{5}$⁠: replace the angle by its supplement, which has the same sine and lies in the principal range.
   Principal range of $\sin^{-1}$⁠: $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ (table, Section 2.2). Lesson: principal values

4. The general rule behind it

   For any $x\in\left[\frac{\pi}{2},\frac{3\pi}{2}\right]$⁠, the angle $\pi-x$ lies in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ and $\sin(\pi-x)=\sin x$⁠. So

   $$\sin^{-1}(\sin x)=\pi-x \qquad\text{for } \frac{\pi}{2}\le x\le\frac{3\pi}{2}.$$

   With $x=\frac{2\pi}{3}$⁠: $\pi-\frac{2\pi}{3}=\frac{\pi}{3}$⁠, as found.

   Seeing the rule makes clear that $\sin^{-1}(\sin x)$ is not $x$ in general: as $x$ runs past $\frac{\pi}{2}$⁠, the value turns round and comes back down. Its graph is a zigzag, not the straight line $y=x$⁠.
   Check the endpoints: $x=\frac{\pi}{2}$ gives $\frac{\pi}{2}$⁠, and $x=\frac{3\pi}{2}$ gives $-\frac{\pi}{2}$⁠, both in the principal range. Lesson: the folding rules

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