Why the Inverse of Sine Only Works on a Chosen Range

Sine takes the same value again and again — $\sin 30°$, $\sin 150°$ and $\sin 390°$ are all $\frac{1}{2}$ — so it is not one-one and has no inverse on all real numbers. Restricting each trigonometric function to a suitable interval makes it bijective, and that restricted piece has an inverse.

This part covers domains, ranges and principal branches, graphs, principal values, and converting between inverse functions.

**Each inverse trigonometric function accepts the values its trigonometric function takes and returns an angle from a fixed principal branch: ${\sin }^{-1}x$ has domain $[-1,1]$ and range $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, ${\cos }^{-1}x$ has domain $[-1,1]$ and range $[0,\pi ]$, and ${\tan }^{-1}x$ has domain $\mathbb{R}$ and range $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$.

All six principal branches:**

- ${\sin }^{-1}x$ — domain $[-1,1]$, range $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$
- ${\cos }^{-1}x$ — domain $[-1,1]$, range $[0,\pi ]$
- ${\tan }^{-1}x$ — domain $\mathbb{R}$, range $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$
- ${\cot }^{-1}x$ — domain $\mathbb{R}$, range $(0,\pi )$
- ${\sec }^{-1}x$ — domain $|x|\ge 1$, range $[0,\pi ]$ without $\frac{\pi }{2}$
- ${\text{cosec}}^{-1}x$ — domain $|x|\ge 1$, range $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ without 0

Worked example. Find the domain of ${\sin }^{-1}(2x-1)$. We need $-1\le 2x-1\le 1$, so $0\le 2x\le 2$ and $0\le x\le 1$.

An everyday example. The inverse sine key on a scientific calculator returns only angles between $-90°$ and $90°$ — exactly the principal branch.

The substance. **${\sin }^{-1}x$ does not mean $\frac{1}{\sin x}$** — the reciprocal is $\text{cosec}x$; here the $-1$ marks an inverse function.

**The graph of each inverse trigonometric function is the reflection, in the line $y=x$, of its trigonometric function restricted to the principal branch, so the roles of the two axes swap.

Key features:**

- $y={\sin }^{-1}x$ — rises from $\left(-1,-\frac{\pi }{2}\right)$ through the origin to $\left(1,\frac{\pi }{2}\right)$
- $y={\cos }^{-1}x$ — falls from $(-1,\pi )$ through $\left(0,\frac{\pi }{2}\right)$ to $(1,0)$
- $y={\tan }^{-1}x$ — rises through the origin towards the horizontal asymptotes $y=\pm \frac{\pi }{2}$
- $y={\cot }^{-1}x$ — falls from near $\pi$ to near 0, passing through $\left(0,\frac{\pi }{2}\right)$
- $y={\sec }^{-1}x$ and $y={\text{cosec}}^{-1}x$ — two separate pieces, defined only for $|x|\ge 1$

Worked example. Plot $y={\sin }^{-1}x$: at $x=-1,-\frac{1}{2},0,\frac{1}{2},1$ the values are $-\frac{\pi }{2},-\frac{\pi }{6},0,\frac{\pi }{6},\frac{\pi }{2}$, or about $-1.57,-0.52,0,0.52,1.57$.

An everyday example. Finding the angle of a ramp from its height and length reads straight off the inverse sine graph — a 1 m rise along a 2 m ramp gives ${\sin }^{-1}0.5=30°$.

The substance. **The horizontal asymptotes of ${\tan }^{-1}x$ come from the vertical asymptotes of $\tan x$** — reflection in $y=x$ turns one kind into the other.

The principal value is the unique angle in the principal branch whose trigonometric value equals the given number, found by locating a reference angle and then choosing the sign or quadrant allowed by the branch.

Negative arguments:

- ${\sin }^{-1}(-x)=-{\sin }^{-1}x$ and ${\tan }^{-1}(-x)=-{\tan }^{-1}x$
- ${\cos }^{-1}(-x)=\pi -{\cos }^{-1}x$ and ${\cot }^{-1}(-x)=\pi -{\cot }^{-1}x$

Worked examples:

- ${\sin }^{-1}\left(-\frac{\sqrt{3}}{2}\right)=-\frac{\pi }{3}$
- ${\cos }^{-1}\left(-\frac{1}{2}\right)=\pi -\frac{\pi }{3}=\frac{2\pi }{3}$
- ${\tan }^{-1}(-1)=-\frac{\pi }{4}$
- ${\sec }^{-1}(-2)={\cos }^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi }{3}$

Worked example 2. Evaluate ${\sin }^{-1}\left(\sin \frac{3\pi }{5}\right)$. The angle $\frac{3\pi }{5}$ lies outside $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, but $\sin \frac{3\pi }{5}=\sin \left(\pi -\frac{3\pi }{5}\right)=\sin \frac{2\pi }{5}$, so the answer is $\frac{2\pi }{5}$.

An everyday example. A navigation display reporting a bearing must pick one angle out of many equivalent ones, just as a principal value picks one angle from the branch.

The substance. **${\sin }^{-1}(\sin x)=x$ only when x lies in the principal branch** — otherwise the angle must first be brought back into the branch.

**To convert, let the given inverse function equal $\theta$, sketch a right-angled triangle with the matching sides, find the third side by Pythagoras, and read off the required ratio — adjusting for negative arguments.

Standard conversions for positive x:**

- ${\sin }^{-1}x={\cos }^{-1}\sqrt{1-x^2}={\tan }^{-1}\frac{x}{\sqrt{1-x^2}}$ for $0<x<1$
- ${\tan }^{-1}x={\sin }^{-1}\frac{x}{\sqrt{1+x^2}}={\cos }^{-1}\frac{1}{\sqrt{1+x^2}}$
- ${\text{cosec}}^{-1}x={\sin }^{-1}\frac{1}{x}$ and ${\sec }^{-1}x={\cos }^{-1}\frac{1}{x}$ for $|x|\ge 1$
- ${\cot }^{-1}x={\tan }^{-1}\frac{1}{x}$ for $x>0$

Worked example. Write ${\sin }^{-1}\frac{3}{5}$ in terms of ${\tan }^{-1}$. If $\theta ={\sin }^{-1}\frac{3}{5}$, the opposite side is 3 and the hypotenuse 5, so the adjacent side is $\sqrt{25-9}=4$. Hence $\theta ={\tan }^{-1}\frac{3}{4}$, about $36.87°$.

Worked example 2. Evaluate $\cos \left({\tan }^{-1}\frac{5}{12}\right)$. The hypotenuse is $\sqrt{144+25}=13$, so the value is $\frac{12}{13}\approx 0.923$.

An everyday example. A roof rising 3 m over a horizontal run of 4 m has pitch ${\tan }^{-1}\frac{3}{4}$ — the same angle as ${\sin }^{-1}\frac{3}{5}$ measured along its 5 m rafter.

The substance. **${\cot }^{-1}x={\tan }^{-1}\frac{1}{x}$ fails for negative x** — for $x<0$ it becomes $\pi +{\tan }^{-1}\frac{1}{x}$.