Composite functions: arcsin of sin x and sin of arcsin x

What you will be able to do

• say when $\sin(\sin^{-1}x) = x$ and when $\sin^{-1}(\sin x) = x$;
• find $\sin^{-1}(\sin x)$, $\cos^{-1}(\cos x)$ and $\tan^{-1}(\tan x)$ for any $x$;
• sketch the three zigzag graphs and read off their periods.

Doing and undoing, in two orders

An inverse function undoes the function. But the two orders are not equal.

Inverse first, then the function. $\sin^{-1}x$ is an angle whose sine is $x$. Taking the sine of that angle gives back $x$:

For $x$ outside $[-1, 1]$ the inner part does not exist. So $\sin(\sin^{-1}2)$ is not defined, even though it looks like it should be $2$. In the same way $\cos(\cos^{-1}x) = x$ for $-1 \le x \le 1$, and $\tan(\tan^{-1}x) = x$ for every real $x$.

The function first, then the inverse. Here the inverse returns the angle in the branch that has the same sine. If $x$ is already in the branch, nothing changes. If not, the answer is a different angle.

Try $x = \dfrac{5\pi}{6}$. Then $\sin x = \dfrac{1}{2}$, and $\sin^{-1}\dfrac{1}{2} = \dfrac{\pi}{6}$. So $\sin^{-1}\left(\sin\dfrac{5\pi}{6}\right) = \dfrac{\pi}{6}$, which is not $\dfrac{5\pi}{6}$.

The three rules, piece by piece

To find $\sin^{-1}(\sin x)$, find the angle in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ with the same sine. Use $\sin(\pi - x) = \sin x$ and the period $2\pi$.

• If $-\dfrac{\pi}{2} \le x \le \dfrac{\pi}{2}$, the result is $x$.
• If $\dfrac{\pi}{2} \le x \le \dfrac{3\pi}{2}$, the result is $\pi - x$.
• If $-\dfrac{3\pi}{2} \le x \le -\dfrac{\pi}{2}$, the result is $-\pi - x$.
• Then the pattern repeats every $2\pi$.

For $\cos^{-1}(\cos x)$ the branch is $[0, \pi]$.

• If $0 \le x \le \pi$, the result is $x$.
• If $\pi \le x \le 2\pi$, the result is $2\pi - x$.
• If $-\pi \le x \le 0$, the result is $-x$, because the cosine is even.
• The pattern repeats every $2\pi$.

For $\tan^{-1}(\tan x)$ the branch is $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$. If $x$ is in it, the result is $x$. If $\dfrac{\pi}{2} < x < \dfrac{3\pi}{2}$, the result is $x - \pi$. The pattern repeats every $\pi$, and the function is not defined at odd multiples of $\dfrac{\pi}{2}$.

A worked example

Find (a) $\sin^{-1}(\sin 4)$, (b) $\cos^{-1}(\cos 5)$, (c) $\tan^{-1}(\tan 2)$, (d) $\sin^{-1}(\sin 10)$. The angles are in radians.

1. (a) $\dfrac{\pi}{2} \approx 1.571$ and $\dfrac{3\pi}{2} \approx 4.712$, so $4$ lies between them. The result is $\pi - 4 \approx -0.858$. Check: $\sin 4 \approx -0.757$ and $\sin(-0.858) \approx -0.757$.
2. (b) $5$ lies between $\pi \approx 3.142$ and $2\pi \approx 6.283$. The result is $2\pi - 5 \approx 1.283$.
3. (c) $2$ lies between $\dfrac{\pi}{2}$ and $\dfrac{3\pi}{2}$. The result is $2 - \pi \approx -1.142$.
4. (d) Take away one period: $10 - 2\pi \approx 3.717$, which lies between $\dfrac{\pi}{2}$ and $\dfrac{3\pi}{2}$. The result is $\pi - 3.717 = 3\pi - 10 \approx -0.575$.

Practise with new numbers

Try it in Python

The program compares $\sin^{-1}(\sin x)$ with the piece-by-piece rule for six values of $x$.

Common mistakes

• Writing $\sin^{-1}(\sin x) = x$ for every $x$. It holds only on $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$.
• Writing $\sin(\sin^{-1}x) = x$ for every $x$. It holds only on $[-1, 1]$.
• A negative answer for $\cos^{-1}(\cos x)$. The result is always in $[0, \pi]$.
• Using the period $2\pi$ for $\tan^{-1}(\tan x)$. The tangent repeats every $\pi$.

Remember these

• $\sin(\sin^{-1}x) = x$ and $\cos(\cos^{-1}x) = x$ for $x$ in $[-1, 1]$. $\tan(\tan^{-1}x) = x$ for all real $x$.
• $\sin^{-1}(\sin x) = x$ only for $x$ in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, and $\cos^{-1}(\cos x) = x$ only for $x$ in $[0, \pi]$.
• Otherwise find the angle in the branch with the same sine, cosine or tangent.
• The zigzag graphs have period $2\pi$ for sine and cosine, and $\pi$ for tangent.

Check yourself

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

1. Find $\sin\left(\sin^{-1}\dfrac{3}{5}\right)$.

2. Find $\sin^{-1}\left(\sin\dfrac{5\pi}{6}\right)$.

3. Find $\cos^{-1}\left(\cos\left(-\dfrac{\pi}{3}\right)\right)$.

4. Find $\tan^{-1}\left(\tan\dfrac{3\pi}{4}\right)$.

5. Is $\sin(\sin^{-1}1.5)$ equal to $1.5$? Explain.

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.