Domains, Ranges and Graphs of Inverse Trigonometric Functions: Quick Revision

A brief note on Domains, Ranges and Graphs of Inverse Trigonometric Functions. Read it before a test. Try each question at the end before you open its answer.

The branches

• Why a cut: the sine, cosine and tangent repeat their values, so each is cut to a piece where it takes every value once.
• $\sin^{-1}$: domain $[-1, 1]$, range $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$.
• $\cos^{-1}$: domain $[-1, 1]$, range $[0, \pi]$.
• $\tan^{-1}$: domain all real numbers, range $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.
• $\cot^{-1}$: domain all real numbers, range $(0, \pi)$.
• $\sec^{-1}$ and $\operatorname{cosec}^{-1}$: domain $|x| \ge 1$. The range of $\sec^{-1}$ is $[0, \pi]$ without $\dfrac{\pi}{2}$. The range of $\operatorname{cosec}^{-1}$ is $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ without $0$.
• Not a reciprocal: $\sin^{-1}x$ is an angle. The reciprocal of $\sin x$ is $\operatorname{cosec} x$.

The graphs

• Mirror: the graph of $f^{-1}$ is the graph of $f$ reflected in $y = x$.
• $\sin^{-1}$ and $\tan^{-1}$: rising and odd. $\tan^{-1}$ has the asymptotes $y = \pm\dfrac{\pi}{2}$.
• $\cos^{-1}$: falling, from $(-1, \pi)$ to $(1, 0)$, through $\left(0, \dfrac{\pi}{2}\right)$.
• Reciprocals: $\sec^{-1}x = \cos^{-1}\dfrac{1}{x}$ and $\operatorname{cosec}^{-1}x = \sin^{-1}\dfrac{1}{x}$ for $|x| \ge 1$.

Composite functions

• Inverse first: $\sin(\sin^{-1}x) = x$ and $\cos(\cos^{-1}x) = x$ for $x$ in $[-1, 1]$. $\tan(\tan^{-1}x) = x$ for all $x$.
• Function first: $\sin^{-1}(\sin x) = x$ only for $x$ in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$. For $x$ in $\left[\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right]$ it is $\pi - x$.
• Cosine: $\cos^{-1}(\cos x) = x$ for $x$ in $[0, \pi]$, and $2\pi - x$ for $x$ in $[\pi, 2\pi]$.
• Tangent: $\tan^{-1}(\tan x) = x - \pi$ for $x$ in $\left(\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right)$. Its period is $\pi$.

A worked example

Find the domain of $\cos^{-1}(3x - 2)$, and the value of $\sin^{-1}(\sin 8)$ with $8$ in radians.

1. The domain. We need $-1 \le 3x - 2 \le 1$, so $1 \le 3x \le 3$ and $\dfrac{1}{3} \le x \le 1$.
2. The value. $8 - 2\pi \approx 1.717$ lies between $\dfrac{\pi}{2}$ and $\dfrac{3\pi}{2}$. So the result is $\pi - (8 - 2\pi) = 3\pi - 8 \approx 1.425$.

Practice with new numbers

A program to try

The program checks the end values of the branches and two composite values.

Mistakes to avoid

• A reciprocal: reading $\sin^{-1}x$ as $\dfrac{1}{\sin x}$.
• A negative inverse cosine: the range of $\cos^{-1}$ is $[0, \pi]$.
• The branch ignored: writing $\sin^{-1}(\sin x) = x$ for every $x$.
• The wrong period: $\tan^{-1}(\tan x)$ repeats every $\pi$, not $2\pi$.

Check yourself

1. State the range of $\cos^{-1}$.

2. Find $\tan^{-1}\left(-\sqrt{3}\right)$.

3. Find $\sec^{-1}(-1)$.

4. Find $\cos^{-1}\left(\cos\dfrac{7\pi}{4}\right)$.

5. Is $\cos(\cos^{-1}2)$ defined?

Sources

• NCERT, Mathematics Textbook for Class XII, Part I, Chapter 2: Inverse Trigonometric Functions.
• OpenStax, Algebra and Trigonometry 2e, Section 8.3: Inverse Trigonometric Functions.