Graphs and the other three inverse functions

What you will be able to do

• draw the graphs of $\sin^{-1}$, $\cos^{-1}$ and $\tan^{-1}$ by reflecting the cut graphs in $y = x$;
• read off the domain, range, symmetry and asymptotes from a graph;
• state the branches of $\cot^{-1}$, $\sec^{-1}$ and $\operatorname{cosec}^{-1}$;
• find principal values of these three.

Reflecting in the line $y = x$

An inverse swaps inputs and outputs. If $(a, b)$ is a point of $f$, then $(b, a)$ is a point of $f^{-1}$. On the graph this is a reflection in the line $y = x$.

For example $\left(\dfrac{\pi}{6}, \dfrac{1}{2}\right)$ is on the sine branch. So $\left(\dfrac{1}{2}, \dfrac{\pi}{6}\right)$ is on the graph of $\sin^{-1}$. The domain of $f$ becomes the range of $f^{-1}$, and the range of $f$ becomes the domain of $f^{-1}$.

The three main graphs

Reflect the three cut graphs, and you get these pictures.

• $y = \sin^{-1}x$ rises from $\left(-1, -\dfrac{\pi}{2}\right)$ to $\left(1, \dfrac{\pi}{2}\right)$ through the origin. It is an odd function, so the graph is unchanged by a half turn about the origin: $\sin^{-1}(-x) = -\sin^{-1}x$.
• $y = \cos^{-1}x$ falls from $(-1, \pi)$ to $(1, 0)$ and crosses the $y$-axis at $\dfrac{\pi}{2}$. It is neither odd nor even. A half turn about the point $\left(0, \dfrac{\pi}{2}\right)$ leaves it unchanged, which says $\cos^{-1}(-x) = \pi - \cos^{-1}x$.
• $y = \tan^{-1}x$ rises through the origin and is odd. It flattens towards the horizontal lines $y = \dfrac{\pi}{2}$ and $y = -\dfrac{\pi}{2}$. These lines are asymptotes: the graph comes as close as you like but never reaches them.

The sine and cosine branches are steep at their ends, so the inverse graphs are steep at $x = \pm 1$. The inverse sine and cosine have a vertical tangent there.

The other three inverse functions

The cotangent, secant and cosecant need cuts too.

• $\cot^{-1}$. The cotangent falls from very large to very small values on $(0, \pi)$. So $\cot^{-1}$ has domain all real numbers and range $(0, \pi)$.
• $\sec^{-1}$. The secant is $\dfrac{1}{\cos x}$, so it is never between $-1$ and $1$. On $[0, \pi]$ it is defined everywhere except at $\dfrac{\pi}{2}$. So $\sec^{-1}$ has domain $(-\infty, -1] \cup [1, \infty)$ and range $[0, \pi]$ without $\dfrac{\pi}{2}$.
• $\operatorname{cosec}^{-1}$. The cosecant is $\dfrac{1}{\sin x}$. On $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ it is defined everywhere except at $0$. So $\operatorname{cosec}^{-1}$ has domain $(-\infty, -1] \cup [1, \infty)$ and range $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ without $0$.

To find a value, ask the question the definition asks. What angle in $[0, \pi]$ has the secant $2$? The cosine of that angle is $\dfrac{1}{2}$, so the angle is $\dfrac{\pi}{3}$. In general, for $|x| \ge 1$,

For the cotangent, $\cot^{-1}1 = \dfrac{\pi}{4}$ and $\cot^{-1}(-1) = \dfrac{3\pi}{4}$, because the range is $(0, \pi)$. The next course shows that $\cot^{-1}x = \dfrac{\pi}{2} - \tan^{-1}x$ for every real $x$.

Where books differ. The ranges above are the ones of the NCERT text. A-Level, IB and AP use only the first three inverse functions. Some books and software take another range for $\cot^{-1}$, such as $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ with $0$ left out, and another for $\sec^{-1}$. If a question depends on it, check which branch the book states.

A worked example

Find (a) $\sec^{-1}(-2)$, (b) $\operatorname{cosec}^{-1}(-2)$, (c) $\cot^{-1}(-\sqrt{3})$.

1. (a) We need $\cos y = -\dfrac{1}{2}$ with $y$ in $[0, \pi]$. So $y = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$.
2. (b) We need $\sin y = -\dfrac{1}{2}$ with $y$ in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$. So $y = -\dfrac{\pi}{6}$.
3. (c) We need $\cot y = -\sqrt{3}$ with $y$ in $(0, \pi)$. The cotangent is negative in the second quadrant, and $\cot\dfrac{\pi}{6} = \sqrt{3}$. So $y = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$.

Practise with new numbers

Try it in Python

The program finds the three inverse functions of a few numbers, using the inverse cosine, sine and tangent that a calculator has.

Common mistakes

• Using the wrong range for $\cot^{-1}$. With the NCERT range $(0, \pi)$, $\cot^{-1}(-1)$ is $\dfrac{3\pi}{4}$, not $-\dfrac{\pi}{4}$.
• Putting $\dfrac{\pi}{2}$ in the range of $\sec^{-1}$. The secant of $\dfrac{\pi}{2}$ does not exist.
• Taking $\sec^{-1}$ of a number between $-1$ and $1$. The secant is never there, so $\sec^{-1}\dfrac{1}{2}$ is not defined.
• Forgetting that $\tan^{-1}$ never reaches $\pm\dfrac{\pi}{2}$. The lines are asymptotes.

Remember these

• The graph of $f^{-1}$ is the graph of $f$ reflected in $y = x$.
• $\tan^{-1}$ has the asymptotes $y = \pm\dfrac{\pi}{2}$. $\sin^{-1}$ and $\cos^{-1}$ have the domain $[-1, 1]$.
• $\cot^{-1}:\mathbb{R} \to (0, \pi)$. $\sec^{-1}$ and $\operatorname{cosec}^{-1}$ have the domain $|x| \ge 1$.
• $\sec^{-1}x = \cos^{-1}\dfrac{1}{x}$ and $\operatorname{cosec}^{-1}x = \sin^{-1}\dfrac{1}{x}$ for $|x| \ge 1$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

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

2. The graph of $y = \tan^{-1}x$ has two horizontal asymptotes. Give their equations.

3. Find $\cot^{-1}\left(-\dfrac{1}{\sqrt{3}}\right)$.

4. Find $\operatorname{cosec}^{-1}\sqrt{2}$.

5. Why is $\dfrac{\pi}{2}$ left out of the range of $\sec^{-1}$?

References

• NCERT, Mathematics Textbook for Class XII, Part I, Chapter 2: Inverse Trigonometric Functions.
• OpenStax, Algebra and Trigonometry 2e, Section 8.3: Inverse Trigonometric Functions.
• OpenStax, Calculus Volume 1, Section 1.4: Inverse Functions.