Inverse Trigonometric Functions: Principal Values (sin⁻¹x, cos⁻¹x, tan⁻¹x)

Introduction

The trigonometric functions $\sin x$ , $\cos x$ , and $\tan x$ are many-to-one — many different inputs give the same output. This means they do not have inverses unless we deliberately restrict their domains to a chosen interval. The resulting functions are called the inverse trigonometric functions, written $\sin^{-1}x$ , $\cos^{-1}x$ , and $\tan^{-1}x$ , and the specific output interval chosen for each is called the principal value range.

In the 9709 exam, questions routinely ask you to evaluate inverse trig expressions, solve equations whose solutions must lie in a specified interval, or identify the range of an inverse trig function — all of which require a precise understanding of principal values and the correct notation.

Important notation warning: $\sin^{-1}x$ does not mean $\dfrac{1}{\sin x}$ . The superscript $-1$ here denotes the inverse function, not a reciprocal. The reciprocal of $\sin x$ is written $\cosec x$ .

Core Concept

Because $\sin$ , $\cos$ , and $\tan$ are periodic and many-to-one, we must restrict each to a one-to-one piece of its graph before an inverse function can exist. The restricted intervals chosen are the conventional principal value domains.

Function	Restricted domain	Principal value range of inverse
$y = \sin x$	$-\dfrac{\pi}{2} \leq x \leq \dfrac{\pi}{2}$	$-\dfrac{\pi}{2} \leq \sin^{-1}x \leq \dfrac{\pi}{2}$
$y = \cos x$	$0 \leq x \leq \pi$	$0 \leq \cos^{-1}x \leq \pi$
$y = \tan x$	$-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}$	$-\dfrac{\pi}{2} < \tan^{-1}x < \dfrac{\pi}{2}$

The domain of each inverse function is the range of the original restricted function:

Inverse function	Domain (valid inputs)
$\sin^{-1}x$	$-1 \leq x \leq 1$
$\cos^{-1}x$	$-1 \leq x \leq 1$
$\tan^{-1}x$	$x \in \mathbb{R}$ (all real numbers)

Graphically, each inverse function is the reflection of its restricted parent curve in the line $y = x$ .

Key Formulae & Definitions

Defining relationship (valid only within the principal value range):

\sin^{-1}x = \theta \iff \sin\theta = x, \quad -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}

\cos^{-1}x = \theta \iff \cos\theta = x, \quad 0 \leq \theta \leq \pi

\tan^{-1}x = \theta \iff \tan\theta = x, \quad -\frac{\pi}{2} < \theta < \frac{\pi}{2}

Cancellation identities (only when the argument is in the principal range):

\sin(\sin^{-1}x) = x \qquad \sin^{-1}(\sin\theta) = \theta \text{ only if } -\tfrac{\pi}{2} \leq \theta \leq \tfrac{\pi}{2}

Useful exact values drawn from standard triangles:

\sin^{-1}\!\left(\tfrac{1}{2}\right) = \frac{\pi}{6}, \quad \cos^{-1}\!\left(\tfrac{1}{2}\right) = \frac{\pi}{3}, \quad \tan^{-1}(1) = \frac{\pi}{4}

\sin^{-1}\!\left(\tfrac{\sqrt{3}}{2}\right) = \frac{\pi}{3}, \quad \cos^{-1}\!\left(-1\right) = \pi, \quad \tan^{-1}\!\left(-\sqrt{3}\right) = -\frac{\pi}{3}

Worked Examples

Example 1 — Evaluating an inverse trig expression exactly

Find the exact value of $\cos^{-1}\!\left(-\dfrac{\sqrt{2}}{2}\right)$ , giving your answer in radians.

Step 1: Let $\theta = \cos^{-1}\!\left(-\dfrac{\sqrt{2}}{2}\right)$ . By definition, $\cos\theta = -\dfrac{\sqrt{2}}{2}$ with $0 \leq \theta \leq \pi$ .

Step 2: Identify the reference angle. We know $\cos\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}$ , so the reference angle is $\dfrac{\pi}{4}$ .

Step 3: Since $\cos\theta$ is negative and $\theta$ must lie in $[0, \pi]$ , the angle is in the second quadrant:

\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}

Answer: $\cos^{-1}\!\left(-\dfrac{\sqrt{2}}{2}\right) = \dfrac{3\pi}{4}$

Example 2 — Applying the principal value correctly in an equation

Solve $\sin^{-1}(2x - 1) = -\dfrac{\pi}{6}$ , giving an exact answer.

Step 1: Apply $\sin$ to both sides (the inverse relationship).

2x - 1 = \sin\!\left(-\frac{\pi}{6}\right)

Step 2: Evaluate $\sin\!\left(-\dfrac{\pi}{6}\right)$ .

\sin\!\left(-\frac{\pi}{6}\right) = -\frac{1}{2}

Step 3: Solve the linear equation.

2x - 1 = -\frac{1}{2} \implies 2x = \frac{1}{2} \implies x = \frac{1}{4}

Step 4 (check domain): The input to $\sin^{-1}$ must satisfy $-1 \leq 2x-1 \leq 1$ , i.e. $0 \leq x \leq 1$ . Since $x = \dfrac{1}{4}$ , this is valid. ✓

Answer: $x = \dfrac{1}{4}$

Example 3 — Evaluating a composite expression

Find the exact value of $\sin\!\left(\cos^{-1}\!\left(\dfrac{3}{5}\right)\right)$ .

Step 1: Let $\alpha = \cos^{-1}\!\left(\dfrac{3}{5}\right)$ , so $\cos\alpha = \dfrac{3}{5}$ with $0 \leq \alpha \leq \pi$ .

Step 2: Use the Pythagorean identity $\sin^2\alpha + \cos^2\alpha = 1$ .

\sin^2\alpha = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}

Step 3: Since $\alpha \in [0, \pi]$ , $\sin\alpha \geq 0$ , so we take the positive root.

\sin\alpha = \frac{4}{5}

Answer: $\sin\!\left(\cos^{-1}\!\left(\dfrac{3}{5}\right)\right) = \dfrac{4}{5}$

Common Mistakes & Examiner Pitfalls

Confusing $\sin^{-1}x$ with $(\sin x)^{-1}$ : Remember, $\sin^{-1}x$ is the inverse function; $(\sin x)^{-1} = \cosec x$ . These are entirely different.
Ignoring the principal value range: If asked for $\sin^{-1}\!\left(-\dfrac{1}{2}\right)$ , a common error is to write $\dfrac{7\pi}{6}$ or $\dfrac{11\pi}{6}$ (from the unit circle). The correct answer is $-\dfrac{\pi}{6}$ , since the range of $\sin^{-1}$ is $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ .
Using degrees when radians are expected: Unless a question explicitly asks for degrees, work in radians. The principal value ranges are defined in radians in the 9709 syllabus.
Misapplying the cancellation identity: $\sin^{-1}(\sin\theta) = \theta$ is only true if $\theta \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ . For example, $\sin^{-1}\!\left(\sin\dfrac{2\pi}{3}\right) \neq \dfrac{2\pi}{3}$ ; the correct value is $\dfrac{\pi}{3}$ .
$\tan^{-1}x$ has open endpoints: The range of $\tan^{-1}x$ is the open interval $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ — it never actually reaches $\pm\dfrac{\pi}{2}$ .

Practice Questions

Q1. Write down the exact value of $\tan^{-1}(-1)$ in radians.

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Let $\theta = \tan^{-1}(-1)$ , so $\tan\theta = -1$ with $-\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}$ .

The reference angle satisfying $\tan\theta = 1$ is $\dfrac{\pi}{4}$ . Since the value is negative and the range includes the interval $\left(-\dfrac{\pi}{2}, 0\right)$ :

\theta = -\frac{\pi}{4}

Q2. Find the exact value of $\cos\!\left(\sin^{-1}\!\left(-\dfrac{5}{13}\right)\right)$ .

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Let $\alpha = \sin^{-1}\!\left(-\dfrac{5}{13}\right)$ , so $\sin\alpha = -\dfrac{5}{13}$ with $-\dfrac{\pi}{2} \leq \alpha \leq \dfrac{\pi}{2}$ .

Using $\cos^2\alpha = 1 - \sin^2\alpha$ :

\cos^2\alpha = 1 - \frac{25}{169} = \frac{144}{169}

Since $\alpha \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ , $\cos\alpha \geq 0$ , so:

\cos\alpha = \frac{12}{13}

Q3. Solve $\cos^{-1}(3x + 1) = \dfrac{2\pi}{3}$ , giving your answer as an exact fraction.

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Apply $\cos$ to both sides:

3x + 1 = \cos\!\left(\frac{2\pi}{3}\right) = -\frac{1}{2}

3x = -\frac{3}{2} \implies x = -\frac{1}{2}

Check domain: input to $\cos^{-1}$ must satisfy $-1 \leq 3x+1 \leq 1$ , i.e. $-\dfrac{2}{3} \leq x \leq 0$ . Since $x = -\dfrac{1}{2} \in \left[-\dfrac{2}{3}, 0\right]$ , valid. ✓

Q4. State whether each of the following is defined. If it is, find the exact principal value; if not, explain why.

(a) $\sin^{-1}(1.5)$ $\quad$ (b) $\cos^{-1}(0)$ $\quad$ (c) $\tan^{-1}(-\sqrt{3})$

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(a) $\sin^{-1}(1.5)$ — Not defined. The domain of $\sin^{-1}$ is $[-1, 1]$ , and $1.5 > 1$ .

(b) $\cos^{-1}(0)$ : we need $\cos\theta = 0$ with $0 \leq \theta \leq \pi$ . This gives $\theta = \dfrac{\pi}{2}$ .

(c) $\tan^{-1}(-\sqrt{3})$ : we need $\tan\theta = -\sqrt{3}$ with $-\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}$ . Reference angle: $\tan\dfrac{\pi}{3} = \sqrt{3}$ , so $\theta = -\dfrac{\pi}{3}$ .

Q5. Given that $\sin^{-1}(\sin\theta) = \theta$ only when $\theta$ is in the principal value range, find $\sin^{-1}\!\left(\sin\dfrac{5\pi}{6}\right)$ .

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$\dfrac{5\pi}{6}$ is outside the principal value range $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ , so we cannot apply the cancellation identity directly.

Instead, evaluate $\sin\dfrac{5\pi}{6}$ first:

\sin\frac{5\pi}{6} = \sin\!\left(\pi - \frac{\pi}{6}\right) = \sin\frac{\pi}{6} = \frac{1}{2}

Now apply the inverse:

\sin^{-1}\!\left(\frac{1}{2}\right) = \frac{\pi}{6}

Connections

Prerequisite — Graphs of Trigonometric Functions: Understanding that $\sin x$ , $\cos x$ , and $\tan x$ are periodic and many-to-one is essential to justify why domain restriction is necessary. The shape of each inverse graph is the reflection of its restricted parent in $y = x$ .
Solving Trigonometric Equations: When solving equations such as $\sin x = k$ over a given interval, $\sin^{-1}k$ gives the principal value first solution; you then use symmetry and periodicity of the original function to find all solutions in the required range.
Calculus (Pure Mathematics 2/3): The inverse trigonometric functions appear in differentiation and integration; their derivatives $\dfrac{d}{dx}\sin^{-1}x = \dfrac{1}{\sqrt{1-x^2}}$ etc. depend entirely on correct understanding of the domain and range established here.