10. Trigonometry & Complex Numbers

Inverse circular functions

Inverse circular (trigonometric) functions are fundamental mappings that restrict the domain of periodic trigonometric functions to ensure bijectivity, enabling unique angular retrievals from given ratios.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

📖 Topic progress

○ Mark as complete

Definition and statement

Principal Value Branch Restriction & Inverse Definition

Let $f: [-\frac{\pi}{2}, \frac{\pi}{2}] \to [-1, 1]$ be defined by $f(x) = \sin x$. Since $f$ is strictly monotonic and continuous, it is bijective and possesses a unique inverse function $f^{-1}: [-1, 1] \to [-\frac{\pi}{2}, \frac{\pi}{2}]$ denoted by $f^{-1}(x) = \sin^{-1} x$, satisfying $\sin(\sin^{-1} x) = x$ for all $x \in [-1, 1]$ and $\sin^{-1}(\sin x) = x$ for all $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

📐 Geometric Interpretation

The graph of an inverse circular function is obtained by reflecting the restricted parent trigonometric curve across the line $y = x$, swapping the roles of abscissa (ratio) and ordinate (angle).

⚙️ Algebraic & Structural Meaning

Establishes an isomorphism between multiplicative/ratio domains and angular codomains, converting transcendental equations into algebraic or trigonometric systems.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Domain Restriction for Bijectivity	`Principal branch (e.g., $[-\frac{\pi}{2}, \frac{\pi}{2}]$ for sine)`	Strictly Mandatory / Necessary	Without restriction, trigonometric functions are many-to-one (periodic), making the inverse a non-function (multivalued relation).
Continuity of the Parent Function	`Closed interval of restricted domain`	Strictly Mandatory	Discontinuities would destroy the intermediate value property and monotonicity, preventing topological homeomorphism.
Strict Monotonicity	`Open interior of restricted domain`	Essential for invertibility	If derivative vanishes identically on an interval, the function is not strictly monotonic and fails the horizontal line test.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Negative Argument Identities (Parity)

$$ \sin^{-1}(-x) = -\sin^{-1} x, \quad \cos^{-1}(-x) = \pi - \cos^{-1} x, \quad \tan^{-1}(-x) = -\tan^{-1} x $$

Rule: Remember that odd functions (\sin, \tan, \csc) pull out the negative sign directly, while even-codomain co-functions (\cos, \sec, \cot) subtract from \pi.

Application: Use this result in the stated type of problem.

Complementary Angle Relations

$$ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}, \quad \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}, \quad \sec^{-1} x + \csc^{-1} x = \frac{\pi}{2} $$

Rule: Whenever matching arguments appear in complementary pairs summing to \frac{\pi}{2}, instantly substitute the constant.

Sum of Two Tangent Inverses

$$ \tan^{-1} x + \tan^{-1} y = \begin{cases} \tan^{-1}\left(\frac{x+y}{1-xy}\right) & \text{if } xy < 1 \\ \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right) & \text{if } x > 0, y > 0, xy > 1 \end{cases} $$

Rule: Always check the product $xy$ before applying the standard formula to avoid missing the $\pi$ phase shift.

Conversion to Sine/Cosine Inverse

$$ \tan^{-1} x = \sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right) = \cos^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) $$

Rule: Draw a mental right triangle with perpendicular $x$ and base $1$ to instantly convert any inverse circular function into another.

Application: Essential time-saver for matching options in multiple-choice questions.

Multiple Angle Inverse Summations

$$ 2\tan^{-1} x = \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) $$

Rule: Use substitution $x = \tan \theta$ to reduce complex algebraic square roots into simple multiple-angle formulas.

Application: Standard pattern in UP PGT calculus and trigonometry integration.

Rapid Elimination via Domain Checks

$$ \text{Domain of } \sec^{-1} x \text{ is } (-\infty, -1] \cup [1, \infty) $$

Rule: Instantly eliminate options where arguments for $\sin^{-1}$ or $\cos^{-1}$ exceed $[-1, 1]$ or where $\sec^{-1}$ lies inside $(-1, 1)$.

Application: Allows candidates to eliminate 2 incorrect options.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

The Out-of-Range Principal Value Fallacy

Common Pitfall: Writing $\sin^{-1}(\sin \frac{2\pi}{3}) = \frac{2\pi}{3}$ without checking if it falls inside $[-\frac{\pi}{2}, \frac{\pi}{2}]\text{.}$

Counterexample: \sin^{-1}\left(\sin \frac{2\pi}{3}\right)

Why it fails: Since $\frac{2\pi}{3} \notin [-\frac{\pi}{2}, \frac{\pi}{2}]$, one must use periodicity: $\sin(\frac{2\pi}{3}) = \sin(\frac{\pi}{3})$, yielding $\frac{\pi}{3}\text{.}$

🛡️ Exam Defense: Always reduce the angle into the exact principal domain interval before applying cancellation property.

⚠️ COMMON ERROR #02

Ignoring Domain Restrictions in Equation Solving

Common Pitfall: Squaring equations or manipulating $\tan^{-1} x + \tan^{-1} y = \tan^{-1} z$ without verifying that the resulting value satisfies the original domain.

Counterexample: \tan^{-1} x + \tan^{-1}(2x) = \frac{\pi}{4}

Why it fails: Extraneous roots often emerge after taking tangents on both sides and clearing denominators.

🛡️ Exam Defense: Substitute obtained values back into the original inverse equation to ensure both sides match.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Sine Inverse	Domain $[-1, 1]$, Range $[-\frac{\pi}{2}, \frac{\pi}{2}]	\sin^{-1}(-x) = -\sin^{-1} x	Symmetric about origin	Odd function symmetry
Cosine Inverse	Domain $[-1, 1]$, Range $[0, \pi]	\cos^{-1}(-x) = \pi - \cos^{-1} x	Decreasing curve from $( -1, \pi )$ to $( 1, 0 )	Offset by $\pi$ for negatives
Tangent Inverse	Domain $\mathbb{R}$, Range $(-\frac{\pi}{2}, \frac{\pi}{2})	\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}	Horizontal asymptotes at $y = \pm \frac{\pi}{2}	All real numbers allowed
Secant Inverse	Domain $(-\infty, -1] \cup [1, \infty)$, Range $[0, \pi] \setminus \{\frac{\pi}{2}\}	\sec^{-1} x = \cos^{-1}(1/x)	Split branches in upper/lower half-planes	Excludes interval $(-1, 1)$

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If $\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{3\pi}{2}$, find the value of $x^2 + y^2 + z^2 - \frac{9}{x+y+z}$.

(A) $0$(B) $1$(C) $-1$(D) $\frac{1}{2}$

📚 Method A: Formal Textbook Derivation

Step 1: Note that the maximum value of $\sin^{-1} x$ is $\frac{\pi}{2}$. Step 2: Since the sum of three principal values equals $\frac{3\pi}{2}$, each must independently achieve its maximum. Step 3: Thus $\sin^{-1} x = \frac{\pi}{2} \implies x = 1$ (and similarly $y=1, z=1$). Step 4: Substitute $x=1, y=1, z=1$ into $1^2 + 1^2 + 1^2 - \frac{9}{1+1+1} = 3 - 3 = 0$.

⚡ Method B: 25-Second Exam Shortcut

Extreme value saturation principle: when sum equals upper bound, each term equals its individual maximum.

💡 Tip: Look out for coefficient multipliers or sums equaling boundary values (like $\pi$ or $\frac{3\pi}{2}$).

CANONICAL PROBLEM 02 Dual-Method Breakdown

The value of $\tan\left(\frac{1}{2} \cos^{-1} \frac{\sqrt{5}}{3}\right)$ is:

(A) $\frac{3-\sqrt{5}}{2}$(B) $\frac{3+\sqrt{5}}{2}$(C) $3 - \sqrt{5}$(D) $\frac{3-\sqrt{5}}{4}$

📚 Method A: Formal Textbook Derivation

Step 1: Let $\theta = \cos^{-1} \frac{\sqrt{5}}{3}$, so $\cos \theta = \frac{\sqrt{5}}{3}$. Step 2: We need $\tan(\theta/2)$. Use half-angle formula $\tan(\theta/2) = \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}$. Step 3: Substitute $\cos \theta = \frac{\sqrt{5}}{3}$ to get $\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}$. Step 4: Rationalize denominator to obtain $\frac{3-\sqrt{5}}{2}$.

⚡ Method B: 25-Second Exam Shortcut

Direct half-angle substitution for inverse trigonometric ratios.

💡 Tip: Rationalizing surd denominators is crucial for matching final option formats.

Practice questions

Choose an answer to check it and see the solution:

The principal value of $\cos^{-1}\left(-\frac{1}{2}\right)$ is:

A $-\frac{\pi}{3}$

B $\frac{2\pi}{3}$

C $\frac{\pi}{3}$

D $-\frac{2\pi}{3}$

Derivation & Analysis:

Since $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$, we have $\pi - \frac{\pi}{3} = \frac{2\pi}{3}$, which lies in $[0, \pi]$.

If $\tan^{-1} 2 + \tan^{-1} 3 + \theta = \pi$, then $\theta$ equals:

A $\frac{\pi}{4}$

B $-\frac{\pi}{4}$

C $\frac{3\pi}{4}$

D 0

Derivation & Analysis:

Compute $\tan^{-1} 2 + \tan^{-1} 3 = \pi + \tan^{-1}\left(\frac{2+3}{1-6}\right) = \pi + \tan^{-1}(-1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$. Thus $\frac{3\pi}{4} + \theta = \pi \implies \theta = \frac{\pi}{4}$.

The domain of the function $f(x) = \sin^{-1}(2x - 1)$ is:

A [-1, 1]

B [0, 1]

C $[-\frac{1}{2}, \frac{1}{2}]$

D (0, 1)

Derivation & Analysis:

For $\sin^{-1}$ argument, $-1 \le 2x - 1 \le 1 \implies 0 \le 2x \le 2 \implies 0 \le x \le 1$.

The value of $\sin\left(\frac{\pi}{3} - \sin^{-1}\left(-\frac{1}{2}\right)\right)$ is:

A 1

B -1

C 0

D $\frac{1}{2}$

Derivation & Analysis:

We know $\sin^{-1}(-1/2) = -\pi/6$. The expression becomes $\sin(\pi/3 - (-\pi/6)) = \sin(\pi/2) = 1$.

If $\cot^{-1} x + \cot^{-1} y + \cot^{-1} z = \frac{\pi}{2}$, then $xy + yz + zx$ is equal to:

A 0

B 1

C -1

D $\frac{1}{2}$

Derivation & Analysis:

Converting to $\tan^{-1}$, $\tan^{-1}(1/x) + \tan^{-1}(1/y) + \tan^{-1}(1/z) = \pi/2 \implies \sum \frac{1}{x} = \frac{1}{x}\frac{1}{y}\frac{1}{z}$ upon taking $\tan$ of both sides, yielding $xy + yz + zx = 1$.

The value of $\tan(\cos^{-1} \frac{3}{5} + \tan^{-1} \frac{1}{4})$ is:

A $\frac{16}{7}$

B $\frac{7}{16}$

C $\frac{19}{8}$

D 1

Derivation & Analysis:

Convert $\cos^{-1}(3/5)$ to $\tan^{-1}(4/3)$. Then evaluate $\tan(\tan^{-1}(4/3) + \tan^{-1}(1/4))$ using $\tan(A+B)$ formula to get $16/7$.

The number of real solutions of equation $\tan^{-1} x + \cot^{-1}(x+1) = \tan^{-1}(1+\frac{1}{x^2})$ is:

A 0

B 1

C 2

D Infinite

Derivation & Analysis:

Simplifying using inverse sum/difference identities yields a unique admissible real solution $x = 1$.

What is the range of $f(x) = \sec^{-1} x$?

A $[-\frac{\pi}{2}, \frac{\pi}{2}]$

B $[0, \pi]$

C $[0, \pi] \setminus \{\frac{\pi}{2}\}$

D $(-\frac{\pi}{2}, \frac{\pi}{2})$

Derivation & Analysis:

By definition, the range of $\sec^{-1} x$ is $[0, \pi]$ excluding $\pi/2$ where cosine is zero.

If $\sin(\sin^{-1} \frac{1}{5} + \cos^{-1} x) = 1$, then the value of $x$ is:

A 1

B $\frac{1}{5}$

C 0

D -1

Derivation & Analysis:

Since $\sin(\theta) = 1 \implies \theta = \pi/2$. Thus $\sin^{-1}(1/5) + \cos^{-1}(x) = \pi/2$, which implies $\cos^{-1}(x) = \cos^{-1}(1/5) \implies x = 1/5$.

Q10

The value of $\cos\left[2 \tan^{-1}\left(\frac{1}{7}\right)\right] + \sin\left[4 \tan^{-1}\left(\frac{1}{3}\right)\right]$ is:

A $\frac{31}{25}$

B 1

C 0

D $\frac{14}{25}$

Derivation & Analysis:

Apply double-angle formulas for inverse tangents: $2\tan^{-1}(1/7) = \cos^{-1}(24/25)$ and compute terms systematically to yield $31/25$.

Common questions

Because inverse functions are strictly bound to their principal ranges. $\tan^{-1}(\tan x) = x$ holds true only when $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$. Outside this interval, periodicity must be accounted for.

Failing to check condition ranges when applying sum formulas like $\tan^{-1} x + \tan^{-1} y = \tan^{-1}(\frac{x+y}{1-xy})$, especially when $xy > 1$ where a $\pi$ shift is mandatory.

Master standard triangle conversions (e.g., converting $\tan^{-1}$ to $\sin^{-1}$ or $\cos^{-1}$ instantly), memorize standard principal branches, and use back-substitution of option values.

← Previous Topic

Height and distance

Next Topic →

General Properties of Inverse Circular Functions