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Inverse Trigonometric Functions
Introduction
Basic Concepts
Practice: Principal Values of Inverse Trig Functions
Properties of Inverse Trigonometric Functions
Practice: Simplifying Inverse Trig Expressions
Miscellaneous Examples
Chapter Review: Mixed Practice
Summary
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Glossary

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Inverse Trigonometric FunctionsBasic Concepts

Reading time: ~50 min

In Class XI you met the six trigonometric functions as functions of a real number. Each one has a domain, the inputs it accepts, and a range, the outputs it can produce:

\sin : \mathbf{R} \to [-1, 1]

\cos : \mathbf{R} \to [-1, 1]

\tan : \mathbf{R} - \left\{ x : x = (2n+1)\dfrac{\pi}{2},\ n \in \mathbf{Z} \right\} \to \mathbf{R}

\cot : \mathbf{R} - \{ x : x = n\pi,\ n \in \mathbf{Z} \} \to \mathbf{R}

\sec : \mathbf{R} - \left\{ x : x = (2n+1)\dfrac{\pi}{2},\ n \in \mathbf{Z} \right\} \to \mathbf{R} - (-1, 1)

\operatorname{cosec} : \mathbf{R} - \{ x : x = n\pi,\ n \in \mathbf{Z} \} \to \mathbf{R} - (-1, 1)

The pieces cut out of a domain are the places where the function is undefined. Four of the six are fractions in disguise, and a fraction breaks down where its denominator is zero. Sort the six functions by where they break down.

tan x
sin x
cosec x
sec x
cos x
cot x
Never breaks down
Breaks where cos x = 0
Breaks where sin x = 0

The tangent and secant have \cos x in the denominator, so they fail at the odd multiples of \dfrac{\pi}{2}; the cotangent and cosecant have \sin x there, so they fail at the multiples of \pi. The odd-looking range \mathbf{R} - (-1, 1) of sec and cosec is the reciprocal at work: since |\cos x| \le 1, the number \dfrac{1}{\cos x} is never smaller than 1 in size, so the whole strip between −1 and 1 is .

Undoing a Function

Recall from Chapter 1: if f : X \to Y with f(x) = y is one-one and onto, there is a function g : Y \to X with g(y) = x. It takes each output of f back to the input it came from, so the domain of g is the range of f, and the range of g is the domain of f. This g is the inverse of f, written f^{-1}; it is also one-one and onto, and its own inverse is f again: (f^{-1})^{-1} = f. As compositions,

(f^{-1} \circ f)(x) = f^{-1}(f(x)) = f^{-1}(y) = x

(f \circ f^{-1})(y) = f(f^{-1}(y)) = f(x) = y

The first line starts from an x in X, the second from a y in Y. Keep that difference in mind: it is why \sin(\sin^{-1} x) and \sin^{-1}(\sin x) will carry different conditions.

Both conditions on f do a job. What goes wrong if f is not one-one?

Some y in Y has no x to go back to
Some y comes from two different inputs, so g(y) would need two answers
f is not defined at some x

Exactly: one-one guarantees at most one way back, and onto guarantees at least one, so together they give exactly one. The sine function takes the value ½ at infinitely many angles, which is why it has no inverse on the whole of \mathbf{R}, as the kite showed.

One Branch at a Time

The sine function has domain \mathbf{R} and range [-1, 1], and it is far from one-one. The way out is to keep only part of its domain: a stretch on which it never repeats a value but still reaches every value from −1 to 1. Such a restriction makes it one-one and onto [-1, 1].

Slide the window. The shaded window picks out one stretch of the sine wave. Find every window on which sine never repeats a value and still reaches every height from −1 to 1.

You have found that the sine function restricted to any of the intervals \left[ -\dfrac{3\pi}{2}, -\dfrac{\pi}{2} \right], \left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right], \left[ \dfrac{\pi}{2}, \dfrac{3\pi}{2} \right] and so on is one-one with range [-1, 1]. On each of them it has an inverse, written \sin^{-1} (the arc sine function), with domain [-1, 1]. Each choice of interval gives a different branch of \sin^{-1}. The branch with range \left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right] is called the principal value branch, and whenever we write \sin^{-1} without saying more, we mean this branch.

For each x \in [-1, 1], \sin^{-1} x is the unique angle y in \left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right] with \sin y = x:

\sin^{-1} : [-1, 1] \to \left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right], \qquad y = \sin^{-1} x \iff \sin y = x

So Aman's kite angle is \sin^{-1}\dfrac{1}{2} = , the one answer from the principal branch. From the definition, \sin(\sin^{-1} x) = x whenever -1 \le x \le 1, and \sin^{-1}(\sin x) = x whenever -\dfrac{\pi}{2} \le x \le \dfrac{\pi}{2}.

Common trap

The two conditions are not the same. \sin(\sin^{-1} x) = x only needs x to be a legal input. But \sin^{-1}(\sin x) = x needs x to be an angle inside the branch already. For example,

\sin^{-1}\left( \sin\dfrac{5\pi}{6} \right) = \sin^{-1}\dfrac{1}{2} = \dfrac{\pi}{6}, not \dfrac{5\pi}{6}.

An answer to a \sin^{-1} question that lies outside \left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right] is always wrong.

The Graph in the Mirror

If y = f(x) is invertible, then x = f^{-1}(y). So whenever (a, b) is a point on the graph of the sine, (b, a) is a point on the graph of \sin^{-1}: the graph of y = \sin^{-1} x is the graph of y = \sin x with the .

Fig 2.1 (i) shows the sine curve and Fig 2.1 (ii) the same curve with its axes swapped, which gives every branch of \sin^{-1}. In both, the green strip marks the principal value branch, drawn thick (the dark portion in your textbook).

Fig 2.1 (i) — y = \sin x

Fig 2.1 (ii) — the sine curve with its axes swapped: y = \sin^{-1} x, all branches

Swapping the two coordinates of a point is exactly what reflecting it in the line y = x does, so the graph of an inverse is the of the original graph in that line. The widget below is Fig 2.1 (iii), both graphs on the same axes. Drag P along the principal branch of the sine; its reflection Q in the dashed line y = x traces the graph of \sin^{-1} x.

The reason is a single line of algebra: b = \sin a says exactly the same thing as a = \sin^{-1} b when a lies in the branch. So the highest point \left( \dfrac{\pi}{2}, 1 \right) of the sine lands on the rightmost point \left( 1, \dfrac{\pi}{2} \right) of the arc sine, and the whole branch flips across the line y = x.

Inverse of the Cosine Function

The cosine function also has domain \mathbf{R} and range [-1, 1]. Restricted to [0, \pi] it falls steadily from 1 to −1, so it becomes one-one and onto with range [-1, 1]. The same is true on [-\pi, 0], [\pi, 2\pi] and so on: on each of these intervals cosine is bijective, and each gives a branch of the inverse, \cos^{-1} (the arc cosine function), with domain [-1, 1]. The branch with range [0, \pi] is the principal value branch:

\cos^{-1} : [-1, 1] \to [0, \pi]

The graph of y = \cos^{-1} x is drawn just like that of y = \sin^{-1} x. In Fig 2.2 the cosine curve and the arc cosine share the green strip [0, \pi].

Fig 2.2 (i) — y = \cos x

Fig 2.2 (ii) — y = \cos^{-1} x

Why does cosine not simply borrow the sine's interval \left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right]?

Cosine is undefined at −π/2 and π/2
cos(−π/3) = cos(π/3), so cosine repeats values there, and it never goes below 0
Cosine is decreasing on that interval

On [0, \pi] cosine slides once from 1 down to −1 without repeating itself, and that is why the principal branch of \cos^{-1} is [0, \pi]. One consequence to remember: \cos^{-1} of a negative number is an angle, never a negative one.

Inverse of the Cosecant Function

Since \operatorname{cosec} x = \dfrac{1}{\sin x}, the cosecant has domain \{ x : x \in \mathbf{R},\ x \ne n\pi,\ n \in \mathbf{Z} \} and range \{ y : y \in \mathbf{R},\ y \ge 1 \text{ or } y \le -1 \}, i.e. \mathbf{R} - (-1, 1). It takes every real value except those strictly between −1 and 1, and it is undefined at the multiples of \pi.

Restricted to \left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right] - \{ 0 \} it is one-one and onto \mathbf{R} - (-1, 1); so it is on \left[ -\dfrac{3\pi}{2}, -\dfrac{\pi}{2} \right] - \{ -\pi \}, \left[ \dfrac{\pi}{2}, \dfrac{3\pi}{2} \right] - \{ \pi \} and so on. Each interval gives a branch of \operatorname{cosec}^{-1}, whose domain is \mathbf{R} - (-1, 1). The principal value branch is

\operatorname{cosec}^{-1} : \mathbf{R} - (-1, 1) \to \left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right] - \{ 0 \}

It is the sine's branch with one point punched out: at 0 the cosecant would divide by zero, which is the dashed line in the middle of the green strip in Fig 2.3.

Fig 2.3 (i) — y = \operatorname{cosec} x

Fig 2.3 (ii) — y = \operatorname{cosec}^{-1} x

Tip

The input set of \operatorname{cosec}^{-1} loses (-1, 1) for the matching reason: \operatorname{cosec} x is never strictly between −1 and 1, so \operatorname{cosec}^{-1}\dfrac{1}{2} asks for an angle that does not .

Inverse of the Secant Function

Since \sec x = \dfrac{1}{\cos x}, the secant has domain \mathbf{R} - \left\{ x : x = (2n+1)\dfrac{\pi}{2},\ n \in \mathbf{Z} \right\} and range \mathbf{R} - (-1, 1): it is undefined at the multiples of \dfrac{\pi}{2}. Restricted to [0, \pi] - \left\{ \dfrac{\pi}{2} \right\} it is one-one and onto \mathbf{R} - (-1, 1), and so it is on [-\pi, 0] - \left\{ -\dfrac{\pi}{2} \right\}, [\pi, 2\pi] - \left\{ \dfrac{3\pi}{2} \right\} and so on. These give the branches of \sec^{-1}, with domain \mathbf{R} - (-1, 1), and the principal value branch is

\sec^{-1} : \mathbf{R} - (-1, 1) \to [0, \pi] - \left\{ \dfrac{\pi}{2} \right\}

This time the cosine's branch [0, \pi] loses its middle point \dfrac{\pi}{2}, where \cos x = 0. Find the dashed line through the green strip in Fig 2.4.

Fig 2.4 (i) — y = \sec x

Fig 2.4 (ii) — y = \sec^{-1} x

Common trap

\sec^{-1} x is not \dfrac{1}{\cos^{-1} x}. The −1 means "undo the secant", not "one divided by". The reciprocal lives inside: \sec^{-1} 2 asks for the angle whose cosine is \dfrac{1}{2}, which is . And no \sec^{-1} answer can ever be \dfrac{\pi}{2}.

Inverse of the Tangent Function

The tangent has domain \left\{ x : x \in \mathbf{R},\ x \ne (2n+1)\dfrac{\pi}{2},\ n \in \mathbf{Z} \right\} and range \mathbf{R}. Restricted to the open interval \left( -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right) it is one-one and onto \mathbf{R}, and so it is on \left( -\dfrac{3\pi}{2}, -\dfrac{\pi}{2} \right), \left( \dfrac{\pi}{2}, \dfrac{3\pi}{2} \right) and so on. These give the branches of \tan^{-1}, with domain \mathbf{R}; the principal value branch is

\tan^{-1} : \mathbf{R} \to \left( -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right)

The brackets are round. The endpoints \pm\dfrac{\pi}{2} are left out because the tangent is undefined there: its graph runs up a asymptote, the dashed lines in Fig 2.5. In return the tangent sweeps through every real number, so \tan^{-1} accepts any input at all.

Fig 2.5 (i) — y = \tan x

Fig 2.5 (ii) — y = \tan^{-1} x

\tan^{-1} is the workhorse of the six. A road sign on a ghat road that reads "gradient 1 in 10" gives the slope; the angle the road makes with the horizontal is \tan^{-1}\dfrac{1}{10}. In coordinate geometry, a line of slope m has inclination \theta = \tan^{-1} m, and mapping apps use \tan^{-1} to turn the gaps between two points into a direction.

Inverse of the Cotangent Function

The cotangent has domain \{ x : x \in \mathbf{R}, x \ne n\pi, n \in \mathbf{Z} \} and range \mathbf{R}. Restricted to (0, \pi) it is bijective with range \mathbf{R}, and so it is on (-\pi, 0), (\pi, 2\pi) and so on. These give the branches of \cot^{-1}, with domain \mathbf{R}, and the principal value branch is

\cot^{-1} : \mathbf{R} \to (0, \pi)

Fig 2.6 (i) — y = \cot x

Fig 2.6 (ii) — y = \cot^{-1} x

Tip

\cot^{-1} behaves like \cos^{-1}, not like \tan^{-1}: its answers run from 0 to \pi, with both ends left open because \cot 0 and \cot \pi are undefined. So a negative input gives an angle: \cot^{-1}\left( -\dfrac{1}{\sqrt{3}} \right) = \dfrac{2\pi}{3}, even though \tan^{-1}(-\sqrt{3}) = -\dfrac{\pi}{3}.

Principal Value Branches at a Glance

The table collects the six inverse trigonometric functions (principal value branches) with their domains and ranges.

FunctionDomainRange (Principal Value Branch)
\sin^{-1}[-1, 1]\left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right]
\cos^{-1}[-1, 1][0, \pi]
\operatorname{cosec}^{-1}\mathbf{R} - (-1, 1)\left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right] - \{ 0 \}
\sec^{-1}\mathbf{R} - (-1, 1)[0, \pi] - \left\{ \dfrac{\pi}{2} \right\}
\tan^{-1}\mathbf{R}\left( -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right)
\cot^{-1}\mathbf{R}(0, \pi)

Six rows look like a lot to learn. The inverse machine shows they are really two pictures. Pick an inverse, then slide the input x along the number line. The green half of the circle is where that inverse is allowed to answer, the green bar is its domain, and the purple ray is its answer.

Tip

The sin family, \sin^{-1}, \operatorname{cosec}^{-1} and \tan^{-1}, answers from the right half of the circle, \left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right], so a negative input gives a negative principal value. The cos family, \cos^{-1}, \sec^{-1} and \cot^{-1}, answers from the upper half, [0, \pi], so a negative input gives an obtuse angle. Then remember what is removed: 0 from the range of \operatorname{cosec}^{-1}, \dfrac{\pi}{2} from that of \sec^{-1}, and both endpoints for \tan^{-1} and \cot^{-1}.

Three Things to Remember

Note

  1. \sin^{-1} x must not be confused with (\sin x)^{-1}. In fact (\sin x)^{-1} = \dfrac{1}{\sin x}, and similarly for the other trigonometric functions.
  2. Whenever no branch of an inverse trigonometric function is mentioned, we mean its principal value branch.
  3. The value of an inverse trigonometric function that lies in the range of its principal value branch is called its principal value.

Quick check: which of these is the same as (\sin x)^{-1}?

sin⁻¹ x
cosec x
sin (1/x)

Right: (\sin x)^{-1} = \dfrac{1}{\sin x} = \operatorname{cosec} x. The little −1 on \sin^{-1} is a label for "inverse function", not a power.

Finding a Principal Value

Every principal-value question takes the same three moves: call the expression y; turn it into an ordinary trigonometric equation; then pick the one angle that solves the equation and lies in the principal value branch.

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

Solution : Let \sin^{-1}\dfrac{1}{\sqrt{2}} = y. Then \sin y = \dfrac{1}{\sqrt{2}}.

The range of the principal value branch of \sin^{-1} is \left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right], and the angle in it with sine \dfrac{1}{\sqrt{2}} is . So the principal value of \sin^{-1}\dfrac{1}{\sqrt{2}} is \dfrac{\pi}{4}.

Example 2 : Find the principal value of \cot^{-1}\left( -\dfrac{1}{\sqrt{3}} \right).

Solution : Let \cot^{-1}\left( -\dfrac{1}{\sqrt{3}} \right) = y. Then

\cot y = -\dfrac{1}{\sqrt{3}} = -\cot\dfrac{\pi}{3} = \cot\left( \pi - \dfrac{\pi}{3} \right) = \cot\dfrac{2\pi}{3}

The range of the principal value branch of \cot^{-1} is (0, \pi), and \dfrac{2\pi}{3} lies in it. Hence the principal value of \cot^{-1}\left( -\dfrac{1}{\sqrt{3}} \right) is .

Why π − π/3 and not −π/3?

-\cot\dfrac{\pi}{3} = \cot\left( -\dfrac{\pi}{3} \right) is also true, but -\dfrac{\pi}{3} is not in (0, \pi). The identity \cot(\pi - \theta) = -\cot\theta is chosen on purpose: it carries a negative value into the second quadrant, exactly where \cot^{-1} may answer. With a negative input, first ask which family the function belongs to, then pick the identity that lands you in its branch.

Speed Round: Principal Values

Name the family, picture its half of the circle, then answer. After each answer you will see the range and the check.

Get five right to move on.

Sina