Why a Calculator Gives Just One Angle for sin Inverse of a Half

Doubling a number is easy to undo — halve it. But $\sin 30^{\circ }=\frac{1}{2}$, $\sin 150^{\circ }=\frac{1}{2}$, and infinitely many other angles have the same sine. **Asking which angle has sine $\frac{1}{2}$** has no single answer until we agree on a range.

This part covers restricted domains and principal value branches, evaluating principal values, and the graphs of inverse trigonometric functions.

Trigonometric functions repeat their values, so they are not one-one on their natural domains; restricting each to an interval where it is one-one and still takes every value makes it a bijection, and the range of the inverse on that interval is its principal value branch.

Why restriction is needed. A function has an inverse only if it is one-one and onto. On $\mathbb{R}$, $\sin x$ takes each value in $[-1,1]$ infinitely often. Restricted to $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, it is one-one and still covers all of $[-1,1]$.

Principal value branches — domain; range:

- **${\sin }^{-1}x$** — $[-1,1]$; $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$
- **${\cos }^{-1}x$** — $[-1,1]$; $[0,\pi ]$
- **${\tan }^{-1}x$** — $\mathbb{R}$; $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$
- **${\text{cosec}}^{-1}x$** — $\mathbb{R}-(-1,1)$; $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]-\{0\}$
- **${\sec }^{-1}x$** — $\mathbb{R}-(-1,1)$; $[0,\pi ]-\left\{\frac{\pi }{2}\right\}$
- **${\cot }^{-1}x$** — $\mathbb{R}$; $(0,\pi )$

Notation warning. ${\sin }^{-1}x$ means the inverse function, not $\frac{1}{\sin x}$.

**Worked example — why ${\cos }^{-1}$ uses $[0,\pi ]$.** On $[0,\pi ]$, $\cos x$ falls steadily from $1$ to $-1$, taking each value once. On $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ it would repeat:

$$\cos \left(-\frac{\pi }{3}\right)=\cos \frac{\pi }{3}=\frac{1}{2}$$

An everyday example. **A ramp rising $1$ m along a sloping length of $2$ m** makes an angle of ${\sin }^{-1}\frac{1}{2}=\frac{\pi }{6}$, or $30^{\circ }$, with the ground — the principal value is the one that fits a real ramp.

The substance. Other branches exist — $\sin x$ on $\left[\frac{\pi }{2},\frac{3\pi }{2}\right]$ also has an inverse — but the principal branch is the standard choice.

To find a principal value, ask which angle inside the principal value branch has the given trigonometric value, and check that the angle really lies in that branch — especially for negative inputs.

Rules for negative inputs:

$${\sin }^{-1}(-x)=-{\sin }^{-1}x{\tan }^{-1}(-x)=-{\tan }^{-1}x{\text{cosec}}^{-1}(-x)=-{\text{cosec}}^{-1}x$$

$${\cos }^{-1}(-x)=\pi -{\cos }^{-1}x{\sec }^{-1}(-x)=\pi -{\sec }^{-1}x{\cot }^{-1}(-x)=\pi -{\cot }^{-1}x$$

Worked example 1. Since $\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}$ and $\frac{\pi }{4}$ lies in $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$,

$${\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi }{4}$$

Worked example 2.

$${\cos }^{-1}\left(-\frac{1}{2}\right)=\pi -{\cos }^{-1}\frac{1}{2}=\pi -\frac{\pi }{3}=\frac{2\pi }{3}$$

Worked example 3. ${\tan }^{-1}(-\sqrt{3})=-{\tan }^{-1}\sqrt{3}=-\frac{\pi }{3}$, and ${\sec }^{-1}(2)=\frac{\pi }{3}$ because $\sec \frac{\pi }{3}=2$ with $\frac{\pi }{3}\in [0,\pi ]$.

Worked example 4 — a combination.

$${\tan }^{-1}(1)+{\cos }^{-1}\left(-\frac{1}{2}\right)+{\sin }^{-1}\left(-\frac{1}{2}\right)=\frac{\pi }{4}+\frac{2\pi }{3}-\frac{\pi }{6}=\frac{3\pi +8\pi -2\pi }{12}=\frac{3\pi }{4}$$

An everyday example. **A surveyor measuring a hillside with $\tan \theta =1$** takes $\theta ={\tan }^{-1}1=45^{\circ }$, not $225^{\circ }$, because only the principal value fits a real slope.

The substance. **${\sin }^{-1}\left(\sin \frac{2\pi }{3}\right)=\frac{\pi }{3}$, not $\frac{2\pi }{3}$**, because $\frac{2\pi }{3}$ lies outside the principal branch of ${\sin }^{-1}$.

**The graph of each inverse trigonometric function is the reflection, in the line $y=x$, of its parent function's graph on the principal branch, so the parent's domain and range swap axes.

Why reflection works.** If $(a,b)$ lies on $y=\sin x$, then $(b,a)$ lies on $y={\sin }^{-1}x$, and swapping coordinates is exactly reflection in $y=x$.

**Worked example — key points for $y={\sin }^{-1}x$.** The restricted sine graph passes through

$$\left(-\frac{\pi }{2},-1\right),(0,0),\left(\frac{\pi }{2},1\right)$$

so the graph of ${\sin }^{-1}x$ passes through

$$\left(-1,-\frac{\pi }{2}\right),(0,0),\left(1,\frac{\pi }{2}\right)$$

Shapes of the graphs:

- **$y={\sin }^{-1}x$** — increasing, from $\left(-1,-\frac{\pi }{2}\right)$ to $\left(1,\frac{\pi }{2}\right)$
- **$y={\cos }^{-1}x$** — decreasing, from $(-1,\pi )$ to $(1,0)$, through $\left(0,\frac{\pi }{2}\right)$
- **$y={\tan }^{-1}x$** — increasing through the origin, approaching $y=\pm \frac{\pi }{2}$ without touching them
- **$y={\cot }^{-1}x$** — decreasing between $y=0$ and $y=\pi$, through $\left(0,\frac{\pi }{2}\right)$
- **$y={\sec }^{-1}x$ and $y={\text{cosec}}^{-1}x$** — two separate pieces, for $x\le -1$ and $x\ge 1$, with a gap for $-1<x<1$

An everyday example. Tracing a graph on butter paper and flipping the sheet over a diagonal fold swaps left-right with up-down — the same swap that turns a graph into its inverse.

The substance. **The vertical asymptotes of $\tan x$ become horizontal asymptotes of ${\tan }^{-1}x$**, because reflection in $y=x$ turns vertical lines into horizontal ones.