Inverse Trigonometric Functions notes: ISC Class 12 Mathematics

1. Principal-value branches

A trigonometric function is many-one, so it is first restricted to a principal-value branch on which it is one-one; the inverse returns angles only from that branch.

$\sin^{-1}$ : domain $[-1, 1]$ , range $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ ; $\cos^{-1}$ : domain $[-1, 1]$ , range $[0, \pi]$ .
$\tan^{-1}$ : domain $\mathbb{R}$ , range $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ ; $\cot^{-1}$ : domain $\mathbb{R}$ , range $(0, \pi)$ .
$\sec^{-1}$ : domain $\mathbb{R} - (-1, 1)$ , range $[0, \pi] - \left\{\dfrac{\pi}{2}\right\}$ ; $\operatorname{cosec}^{-1}$ : same domain, range $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - \{0\}$ .

$\sin^{-1} x$ is not $\dfrac{1}{\sin x}$ .

Learn this table: almost every principal-value question uses it.
Function	Domain	Principal-value range
$\sin^{-1} x$	$[-1, 1]$	$\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$
$\cos^{-1} x$	$[-1, 1]$	$[0, \pi]$
$\tan^{-1} x$	$\mathbb{R}$	$\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$
$\cot^{-1} x$	$\mathbb{R}$	$(0, \pi)$
$\sec^{-1} x$	$\mathbb{R} - (-1, 1)$	$[0, \pi] - \left\{\dfrac{\pi}{2}\right\}$
$\operatorname{cosec}^{-1} x$	$\mathbb{R} - (-1, 1)$	$\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - \{0\}$

Each value is checked numerically when the chapter is built.
Expression	Branch	Principal value
$\sin^{-1}\left(-\frac12\right)$	sine	$-\dfrac{\pi}{6}$
$\cos^{-1}\frac{\sqrt3}{2}$	cosine	$\dfrac{\pi}{6}$
$\tan^{-1}(-1)$	tangent	$-\dfrac{\pi}{4}$
$\cot^{-1}\left(-\frac{1}{\sqrt3}\right)$	cotangent	$\dfrac{2\pi}{3}$
$\sec^{-1}\sqrt2$	secant	$\dfrac{\pi}{4}$

Formulas

Definition of the inverse sine
$\sin^{-1} x = \theta \iff \sin\theta = x,\ \theta \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$

Key terms

Principal value: The one value of an inverse trigonometric function that lies in its principal range.
Principal-value branch: The restricted interval on which a trigonometric function is one-one and is inverted.

2. Graphs

Each graph is the reflection of its branch in $y = x$ .

$\sin^{-1} x$ rises from $\left(-1, -\dfrac{\pi}{2}\right)$ through $(0, 0)$ to $\left(1, \dfrac{\pi}{2}\right)$ (odd).
$\cos^{-1} x$ falls from $(-1, \pi)$ through $\left(0, \dfrac{\pi}{2}\right)$ to $(1, 0)$ .
$\tan^{-1} x$ rises between the asymptotes $y = \pm\dfrac{\pi}{2}$ ; $\cot^{-1} x$ falls between $y = \pi$ and $y = 0$ .
$\sec^{-1} x$ has two rising pieces and never reaches $\dfrac{\pi}{2}$ .

Transformations scale and shift the range: $y = 3\cos^{-1} x - \pi$ lies in $[-\pi, 2\pi]$ .

$y = \sin^{-1} x$

Domain $[-1, 1]$, range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$; an odd, increasing function.

$y = \cos^{-1} x$

Domain $[-1, 1]$, range $[0, \pi]$; a decreasing function.

$y = \tan^{-1} x$
$y = \cot^{-1} x$

The two curves add up to $\frac{\pi}{2}$ at every $x$.

The inverse sine always answers inside $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
$\theta$	$\sin\theta$	$\sin^{-1}(\sin\theta)$	Same as the angle?
$\dfrac{\pi}{3}$	$\dfrac{\sqrt3}{2}$	$\dfrac{\pi}{3}$	yes
$\dfrac{5\pi}{6}$	$\dfrac12$	$\dfrac{\pi}{6}$	no
$\dfrac{7\pi}{6}$	$-\dfrac12$	$-\dfrac{\pi}{6}$	no
$-\dfrac{3\pi}{4}$	$-\dfrac{1}{\sqrt2}$	$-\dfrac{\pi}{4}$	no

Formulas

Inverse of a trigonometric value
$\sin^{-1}(\sin\theta) = \theta \text{ only for } \theta \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$

Key terms

Horizontal asymptote: A line such as $y = \frac{\pi}{2}$ that the graph of $\tan^{-1} x$ approaches but never meets.

3. Elementary properties

Negative arguments: $\sin^{-1}(-x) = -\sin^{-1} x$ , $\tan^{-1}(-x) = -\tan^{-1} x$ , $\operatorname{cosec}^{-1}(-x) = -\operatorname{cosec}^{-1} x$ ; $\cos^{-1}(-x) = \pi - \cos^{-1} x$ , $\cot^{-1}(-x) = \pi - \cot^{-1} x$ , $\sec^{-1}(-x) = \pi - \sec^{-1} x$ .
Reciprocal relations: $\sin^{-1}\dfrac1x = \operatorname{cosec}^{-1} x$ and $\cos^{-1}\dfrac1x = \sec^{-1} x$ for $|x| \ge 1$ ; $\tan^{-1}\dfrac1x = \cot^{-1} x$ for $x > 0$ and $\cot^{-1} x - \pi$ for $x < 0$ .
Complementary sums: $\sin^{-1} x + \cos^{-1} x = \tan^{-1} x + \cot^{-1} x = \sec^{-1} x + \operatorname{cosec}^{-1} x = \dfrac{\pi}{2}$ .
Interconversion: draw the right triangle; for example $\sin^{-1} x = \tan^{-1}\dfrac{x}{\sqrt{1 - x^2}}$ for $|x| < 1$ .

Odd for sine, tangent and cosecant; "pi minus" for cosine, cotangent and secant.
Rule	Check at $x = 0.6$
$\sin^{-1}(-x) = -\sin^{-1} x$	holds
$\tan^{-1}(-x) = -\tan^{-1} x$	holds
$\cos^{-1}(-x) = \pi - \cos^{-1} x$	holds
$\cot^{-1}(-x) = \pi - \cot^{-1} x$	holds

Use these to change one inverse function into another.
Identity	Valid for	Check
$\sin^{-1} x + \cos^{-1} x = \dfrac{\pi}{2}$	$\|x\| \le 1$	$x = 0.3$ : holds
$\tan^{-1} x + \cot^{-1} x = \dfrac{\pi}{2}$	all real $x$	$x = -4$ : holds
$\sec^{-1} x + \operatorname{cosec}^{-1} x = \dfrac{\pi}{2}$	$\|x\| \ge 1$	$x = 2.5$ : holds

For $0 < x \le 1$: $\cos^{-1} x = \sin^{-1}\sqrt{1 - x^2} = \tan^{-1}\frac{\sqrt{1 - x^2}}{x}$.
Same angle written with	Expression at $x = 0.6$	Value (radians)
$\cos^{-1}$	$\cos^{-1} 0.6$	0.9273
$\sin^{-1}$	$\sin^{-1}\sqrt{1 - 0.36} = \sin^{-1} 0.8$	0.9273
$\tan^{-1}$	$\tan^{-1}\dfrac{0.8}{0.6} = \tan^{-1}\dfrac43$	0.9273
$\sec^{-1}$	$\sec^{-1}\dfrac{1}{0.6} = \sec^{-1}\dfrac53$	0.9273

Formulas

Negative argument for the inverse cosine
$\cos^{-1}(-x) = \pi - \cos^{-1} x$
Complementary identity
$\sin^{-1} x + \cos^{-1} x = \tfrac{\pi}{2}$

Key terms

Interconversion: Writing the same angle with a different inverse trigonometric function, using a right triangle.

4. Sum and difference formulae

$\tan^{-1} x + \tan^{-1} y = \tan^{-1}\dfrac{x + y}{1 - xy}$ if $xy < 1$ ; $= \pi + \tan^{-1}\dfrac{x + y}{1 - xy}$ if $xy > 1$ , $x, y > 0$ ; $= -\pi + \tan^{-1}\dfrac{x + y}{1 - xy}$ if $xy > 1$ , $x, y < 0$ .
$\tan^{-1} x - \tan^{-1} y = \tan^{-1}\dfrac{x - y}{1 + xy}$ if $xy > -1$ .
For $\sin^{-1}$ and $\cos^{-1}$ : name the angles $A$ and $B$ , expand $\sin(A \pm B)$ or $\cos(A \pm B)$ from right triangles, and check that the result lies in the principal range.

Always check the product before using the formula.
Case	$\tan^{-1} x + \tan^{-1} y$ equals
$xy < 1$	$\tan^{-1}\dfrac{x + y}{1 - xy}$
$xy > 1$ , $x, y > 0$	$\pi + \tan^{-1}\dfrac{x + y}{1 - xy}$
$xy > 1$ , $x, y < 0$	$-\pi + \tan^{-1}\dfrac{x + y}{1 - xy}$
Difference, $xy > -1$	$\tan^{-1} x - \tan^{-1} y = \tan^{-1}\dfrac{x - y}{1 + xy}$

In practice, name the angles $A$ and $B$, expand $\sin(A \pm B)$ or $\cos(A + B)$, and check the range.
Formula	Used for
$\sin^{-1} x + \sin^{-1} y = \sin^{-1}\left(x\sqrt{1 - y^2} + y\sqrt{1 - x^2}\right)$	$x, y \ge 0$ , $x^2 + y^2 \le 1$
$\cos^{-1} x + \cos^{-1} y = \cos^{-1}\left(xy - \sqrt{1 - x^2}\sqrt{1 - y^2}\right)$	$x, y \in [-1, 1]$ , $x + y \ge 0$
$\sin^{-1} x - \sin^{-1} y = \sin^{-1}\left(x\sqrt{1 - y^2} - y\sqrt{1 - x^2}\right)$	$x, y \ge 0$ , $x^2 + y^2 \le 1$

Two awkward angles that add to exactly half a right angle.
Step	Working
Check the product	$\dfrac34 \cdot \dfrac17 = \dfrac{3}{28} < 1$
Apply the formula	$\tan^{-1}\dfrac{\frac34 + \frac17}{1 - \frac{3}{28}} = \tan^{-1}\dfrac{\frac{25}{28}}{\frac{25}{28}}$
Answer	$\tan^{-1} 1 = \dfrac{\pi}{4}$

Formulas

Inverse tangent addition
$\tan^{-1} x + \tan^{-1} y = \tan^{-1}\frac{x + y}{1 - xy},\ xy < 1$
Inverse tangent subtraction
$\tan^{-1} x - \tan^{-1} y = \tan^{-1}\frac{x - y}{1 + xy},\ xy > -1$

Key terms

Condition of a formula: The restriction (such as xy < 1) that keeps the combined angle inside the principal range.

5. Double and triple forms

$2\tan^{-1} x = \tan^{-1}\dfrac{2x}{1 - x^2}$ ( $|x| < 1$ ) $= \sin^{-1}\dfrac{2x}{1 + x^2}$ ( $|x| \le 1$ ) $= \cos^{-1}\dfrac{1 - x^2}{1 + x^2}$ ( $x \ge 0$ ).
$2\sin^{-1} x = \sin^{-1}\left(2x\sqrt{1 - x^2}\right)$ for $|x| \le \dfrac{1}{\sqrt2}$ ; $2\cos^{-1} x = \cos^{-1}(2x^2 - 1)$ for $0 \le x \le 1$ .
$3\tan^{-1} x = \tan^{-1}\dfrac{3x - x^3}{1 - 3x^2}$ ( $|x| < \dfrac{1}{\sqrt3}$ ); $3\sin^{-1} x = \sin^{-1}(3x - 4x^3)$ ( $|x| \le \dfrac12$ ); $3\cos^{-1} x = \cos^{-1}(4x^3 - 3x)$ ( $\dfrac12 \le x \le 1$ ).

The conditions keep $2\tan^{-1} x$ inside each principal branch.
$2\tan^{-1} x$ equals	Condition
$\tan^{-1}\dfrac{2x}{1 - x^2}$	$\|x\| < 1$
$\sin^{-1}\dfrac{2x}{1 + x^2}$	$\|x\| \le 1$
$\cos^{-1}\dfrac{1 - x^2}{1 + x^2}$	$0 \le x$

$y = 2\tan^{-1} x$
$y = \tan^{-1}\frac{2x}{1 - x^2}$

For $|x| < 1$ the two curves are the same; outside it they differ by $\pm\pi$.

They come from $\tan 3\theta$, $\sin 3\theta$ and $\cos 3\theta$.
Formula	Condition
$3\tan^{-1} x = \tan^{-1}\dfrac{3x - x^3}{1 - 3x^2}$	$\|x\| < \dfrac{1}{\sqrt3}$
$3\sin^{-1} x = \sin^{-1}(3x - 4x^3)$	$\|x\| \le \dfrac12$
$3\cos^{-1} x = \cos^{-1}(4x^3 - 3x)$	$\dfrac12 \le x \le 1$

Formulas

Double form of the inverse tangent
$2\tan^{-1} x = \tan^{-1}\frac{2x}{1 - x^2},\ |x| < 1$
Triple form of the inverse tangent
$3\tan^{-1} x = \tan^{-1}\frac{3x - x^3}{1 - 3x^2},\ |x| < \tfrac{1}{\sqrt3}$

Key terms

Double form: An expression for $2\tan^{-1} x$ (or $2\sin^{-1} x$ , $2\cos^{-1} x$ ) as a single inverse function.

6. Solving equations

Combine the inverse terms with a sum, double or complementary formula.
Take $\tan$ , $\sin$ or $\cos$ of both sides.
Solve the algebraic equation.
Substitute every root into the original equation; reject roots that break a condition or give a value outside a principal range.

Example: $2\tan^{-1} x = \sin^{-1}\dfrac45$ gives $2x^2 - 5x + 2 = 0$ , roots $\dfrac12$ and $2$ ; only $\dfrac12$ survives.

The last two steps earn the final marks.

Taking sines can add a root; test each one.
Root	$2\tan^{-1} x$	$\sin^{-1}\frac45$	Accept?
$\dfrac12$	0.9273	0.9273	yes
$2$	2.2143	0.9273	no

Formulas

Replacing one inverse function by another
$\sin^{-1} x + \cos^{-1} x = \tfrac{\pi}{2} \Rightarrow \sin^{-1} x = \tfrac{\pi}{2} - \cos^{-1} x$

Key terms

Extraneous root: A root of the algebraic equation that does not satisfy the original inverse-trigonometric equation.

Exam tips

How ISC marks are awarded

Principal value (1 mark): only a value in the correct branch earns the mark.
Identity proofs (3 marks): angles named, formula with its condition, simplification.
Equations (3–5 marks): combining, algebra, testing each root and rejecting with a reason.
Section D (5 marks): questions come in parts such as 2 + 3; label each part.

Common mistakes

Reading $\sin^{-1} x$ as $\dfrac{1}{\sin x}$ .
Cancelling $\sin^{-1}(\sin\theta)$ to $\theta$ when $\theta$ is outside the branch.
Using the inverse tangent addition formula when $xy > 1$ .
Giving a negative value for $\cos^{-1}$ , $\cot^{-1}$ or $\sec^{-1}$ .
Keeping both roots of the quadratic without checking them.

1. Principal-value branches

Formulas

Key terms

2. Graphs

Formulas

Key terms

3. Elementary properties

Formulas

Key terms

4. Sum and difference formulae

Formulas

Key terms

5. Double and triple forms

Formulas

Key terms

6. Solving equations

Formulas

Key terms

Exam tips

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