Inverse sine, inverse cosine and inverse tangent

What you will be able to do

• explain why the sine, cosine and tangent need a cut before they can be undone;
• state the domain and the range of $\sin^{-1}$, $\cos^{-1}$ and $\tan^{-1}$;
• find principal values such as $\sin^{-1}\left(-\dfrac{1}{2}\right)$ and $\cos^{-1}\left(-\dfrac{\sqrt{3}}{2}\right)$ without a calculator;
• read $\sin^{-1}x$ correctly: it is an angle, not $\dfrac{1}{\sin x}$.

An angle from its sine

A ramp rises $1\ \text{m}$ along a slope of $2\ \text{m}$. The sine of its angle is $\dfrac{1}{2}$. What is the angle?

You may say $30^\circ$, which is $\dfrac{\pi}{6}$ radians. But $150^\circ$ has the same sine, and so do $390^\circ$ and $-210^\circ$. The picture shows the dashed line at height $\dfrac{1}{2}$ meeting the sine curve again and again.

Going from an angle to its sine is a function, because each angle has one sine. Going back is not, because one sine belongs to many angles. A rule that gives many answers for one input is not a function.

Making an inverse

A function $f$ has an inverse when different inputs always give different outputs. We say that $f$ is one-one. The inverse sends each output back to the input it came from. Its graph is the graph of $f$ reflected in the line $y = x$.

The sine on all the real numbers is not one-one. So we keep only a piece of its graph where it is one-one and still reaches every value from $-1$ to $1$. The piece from $-\dfrac{\pi}{2}$ to $\dfrac{\pi}{2}$ does this: the sine rises from $-1$ to $1$ and never repeats a value.

The inverse sine, written $\sin^{-1}x$ or $\arcsin x$, is defined like this:

Its domain is $[-1, 1]$ and its range is $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$. The name arcsin comes from the arc: for a unit circle the angle in radians is the length of the arc.

The angle in the chosen piece is called the principal value, and the piece is the principal value branch. Other pieces would also work, for example $\left[\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right]$. But every book, every calculator and every exam uses the branch above, so we use it too.

Cosine and tangent

The cosine on $[0, \pi]$ falls from $1$ to $-1$ and takes every value once. This is the branch for $\cos^{-1}$. The piece from $-\dfrac{\pi}{2}$ to $\dfrac{\pi}{2}$ would not work, because $\cos(-x) = \cos x$ makes every value appear twice.

The tangent on $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ rises from very large negative numbers to very large positive numbers. It never reaches $\pm\dfrac{\pi}{2}$, so the interval is open at both ends. This is the branch for $\tan^{-1}$.

• $\sin^{-1}$: 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 $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.

A-Level, IB and AP use exactly these three, and so does a calculator in radian mode.

Do not confuse $\sin^{-1}x$ with $\dfrac{1}{\sin x}$

The small $-1$ in $\sin^{-1}x$ marks the inverse function, as in $f^{-1}$. It is not a power. The reciprocal of the sine is written $(\sin x)^{-1}$ or $\operatorname{cosec} x$. This is why many books write $\arcsin x$. A power of an inverse function is written with brackets, for example $(\sin^{-1}x)^2$.

Finding a principal value

Follow three steps.

1. Ask the question. Which angle has this sine (or cosine, or tangent)?
2. Use the standard angles. Try $0, \dfrac{\pi}{6}, \dfrac{\pi}{4}, \dfrac{\pi}{3}, \dfrac{\pi}{2}$ first.
3. Put the angle in the branch. For $\sin^{-1}$ and $\tan^{-1}$ a negative number gives a negative angle. For $\cos^{-1}$ a negative number gives an angle between $\dfrac{\pi}{2}$ and $\pi$.

A worked example

Find the principal value of (a) $\sin^{-1}\left(-\dfrac{1}{2}\right)$, (b) $\cos^{-1}\left(-\dfrac{\sqrt{3}}{2}\right)$, (c) $\tan^{-1}\left(-\sqrt{3}\right)$, and say what happens with (d) $\sin^{-1}\sqrt{2}$.

1. (a) $\sin\dfrac{\pi}{6} = \dfrac{1}{2}$ and the sine is odd. So $\sin^{-1}\left(-\dfrac{1}{2}\right) = -\dfrac{\pi}{6}$, which lies in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$.
2. (b) $\cos\dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}$. Since $\cos(\pi - \theta) = -\cos\theta$, the angle is $\pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$, which lies in $[0, \pi]$.
3. (c) $\tan\dfrac{\pi}{3} = \sqrt{3}$ and the tangent is odd. So the angle is $-\dfrac{\pi}{3}$.
4. (d) $\sqrt{2} > 1$, so no angle has this sine. The value is not defined.

Practise with new numbers

Try it in Python

The program lists four angles that all have the sine $\dfrac{1}{2}$. Then it asks the inverse sine for the angle, and gets only one.

Common mistakes

• Reading $\sin^{-1}x$ as $\dfrac{1}{\sin x}$. The first is an angle. The second is the cosecant.
• Giving an angle outside the branch. $\sin^{-1}\dfrac{1}{2}$ is $\dfrac{\pi}{6}$ only. It is not $\dfrac{5\pi}{6}$, although that angle has the same sine.
• A negative inverse cosine. The range of $\cos^{-1}$ is $[0, \pi]$, so its values are never negative.
• A number outside $[-1, 1]$. There is no angle with sine $2$ or cosine $-3$.

Remember these

• $\sin^{-1}:[-1, 1] \to \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ and $\cos^{-1}:[-1, 1] \to [0, \pi]$.
• $\tan^{-1}:\mathbb{R} \to \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.
• $y = \sin^{-1}x$ means $\sin y = x$ with $y$ in the branch.
• $\sin^{-1}x$ is an angle, not $\dfrac{1}{\sin x}$.

Check yourself

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

1. Find the principal value of $\sin^{-1}\dfrac{\sqrt{3}}{2}$.

2. Find the principal value of $\cos^{-1}\left(-\dfrac{\sqrt{2}}{2}\right)$.

3. Find the principal value of $\tan^{-1}(-1)$.

4. Is $\sin^{-1}2$ defined? Give a reason.

5. A student writes $\sin^{-1}x = \dfrac{1}{\sin x}$. Which function is $\dfrac{1}{\sin x}$, and what does $\sin^{-1}x$ mean?

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.