Inverse Trigonometric Functions — chapter notes

A function has an inverse only if it is one-one (each output from exactly one input) and onto its target set. Trigonometric functions are periodic, so every value is taken infinitely often: $\sin\dfrac{\pi}{6} = \sin\dfrac{5\pi}{6} = \sin\dfrac{13\pi}{6} = \dfrac12$. A horizontal line that cuts a graph more than once shows that the function is not one-one.

The fix is to keep one piece of the graph on which the function is one-one and still takes all its values. That piece defines the principal value branch, and the value returned is the principal value. Other branches exist (for example $\left[\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right]$ for sine), but unless stated otherwise we always use the principal branch.

On $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, $\sin x$ increases from $-1$ to $1$. Its inverse is $\sin^{-1}$ (also $\arcsin$):

$y = \sin^{-1} x$ means $\sin y = x$ and $-\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}$.

• Domain $[-1, 1]$; range $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$.
• Negative input gives a negative angle: $\sin^{-1}\left(-\dfrac12\right) = -\dfrac{\pi}{6}$.
• $\sin^{-1} x \ne \dfrac{1}{\sin x}$; the reciprocal is $(\sin x)^{-1} = \operatorname{cosec} x$.

Cosine is one-one on $[0, \pi]$, falling from $1$ to $-1$. So $y = \cos^{-1} x$ means $\cos y = x$ and $0 \le y \le \pi$. Domain $[-1, 1]$; range $[0, \pi]$. Values are never negative; a negative input gives an obtuse angle: $\cos^{-1}\left(-\dfrac12\right) = \dfrac{2\pi}{3}$.

Tangent is one-one on the open interval $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ and takes every real value there. So $y = \tan^{-1} x$ means $\tan y = x$ and $-\dfrac{\pi}{2} < y < \dfrac{\pi}{2}$. Domain $\mathbb{R}$; range the open interval $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$. Example: $\tan^{-1}(-\sqrt3) = -\dfrac{\pi}{3}$.

• $\operatorname{cosec}^{-1} x$: domain $\mathbb{R} - (-1, 1)$ (that is, $|x| \ge 1$); range $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - \{0\}$.
• $\sec^{-1} x$: domain $\mathbb{R} - (-1, 1)$; range $[0, \pi] - \left\{\dfrac{\pi}{2}\right\}$.
• $\cot^{-1} x$: domain $\mathbb{R}$; range $(0, \pi)$.

To find such a value, change to the reciprocal ratio and use the range: $\sec y = -2$ means $\cos y = -\dfrac12$ with $y \in [0, \pi]$, so $\sec^{-1}(-2) = \dfrac{2\pi}{3}$. Note that $\cot^{-1}$ uses $(0, \pi)$, not the tangent range: $\cot^{-1}(-1) = \dfrac{3\pi}{4}$.

Principal value of an inverse function at $a$:

1. Check that $a$ is in the domain (otherwise "not defined").
2. Write the principal range.
3. Find the reference angle whose ratio is $|a|$.
4. Use the sign of $a$ to place the angle in the range.
5. Check: correct ratio and inside the range.

Combinations such as $2\cos^{-1}(0) - 3\tan^{-1}(1) + \sin^{-1}\left(\dfrac12\right)$: find each principal value separately, then combine: $\pi - \dfrac{3\pi}{4} + \dfrac{\pi}{6} = \dfrac{5\pi}{12}$.

Domains: for $\sin^{-1}(u)$ or $\cos^{-1}(u)$ solve $-1 \le u \le 1$. Example: $\cos^{-1}(3 - 2x)$ needs $1 \le x \le 2$. With several terms, intersect the conditions.

The graph of an inverse function is the reflection of the restricted graph in the line $y = x$, because each point $(a, b)$ becomes $(b, a)$.

Uses: compare values without a calculator ($\sin^{-1} 0.8 > \sin^{-1} 0.6$ because the graph rises; $\cos^{-1} 0.8 < \cos^{-1} 0.6$ because it falls); solve inequalities such as $\cos^{-1} x > \dfrac{\pi}{3} \iff -1 \le x < \dfrac12$.

Inside-out, $\sin^{-1}(\sin\theta)$: equals $\theta$ only if $\theta \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$. Otherwise evaluate $\sin\theta$ first and take the principal value: $\sin^{-1}\left(\sin\dfrac{5\pi}{6}\right) = \sin^{-1}\dfrac12 = \dfrac{\pi}{6}$. Likewise $\cos^{-1}(\cos\theta) = \theta$ only on $[0, \pi]$, and $\tan^{-1}(\tan\theta) = \theta$ only on $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.

Outside-in, $\cos\left(\tan^{-1}\dfrac34\right)$: name $\theta = \tan^{-1}\dfrac34$, place it by its range (acute here), draw a right triangle $3$–$4$–$5$: $\cos\theta = \dfrac45$. For a negative input, decide the sign from the range: $\sin\left(\cos^{-1}\left(-\dfrac{5}{13}\right)\right) = \dfrac{12}{13}$ (positive, second quadrant).

Equations: $\tan^{-1}(3x) = -\dfrac{\pi}{3}$ gives $3x = -\sqrt3$, $x = -\dfrac{1}{\sqrt3}$. First check that the right side lies in the range: $\cos^{-1} x = -\dfrac{\pi}{3}$ has no solution.