Trigonometry

Trigonometric ratios of factors and multiples

This section covers the trigonometric ratios of sub-multiple angles (factors like $\frac{\theta}{2}, \frac{\theta}{3}$) and multiple angles ($2\theta, 3\theta, n\theta$).

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

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Definition and statement

Multiple and Sub-Multiple Angle Expansion Principle

For any angle $\theta \in \mathbb{R}$, the trigonometric ratios of multiple angles $n\theta$ ($n \in \mathbb{Z}$) and factor angles $\frac{\theta}{n}$ are uniquely determined by polynomial expansions in terms of $\sin\theta$ and $\cos\theta$, governed by De Moivre's Theorem and fundamental trigonometric identities.

📐 Geometric Interpretation

Sub-multiple angles correspond to chord bisection and angle subdivision on the unit circle $\cos^2\theta + \sin^2\theta = 1$, scaling the frequency of rotation while preserving circular symmetry.

⚙️ Algebraic & Structural Meaning

Provides algebraic links between linear and higher-degree polynomial equations in sine and cosine, enabling root extraction for special angles (e.g., $18^circ, 36^circ, 72^circ$).y

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Domain Compatibility	`\theta \in \mathbb{R} \setminus \{ \frac{\pi}{2} + k\pi \}`	Strictly Mandatory	Division by zero occurs in expressions involving $\tan\theta$ or $\sec\theta$.
Quadrant Restrictions for Radicals	`\text{Sign determination based on quadrant of } \frac{\theta}{2}`	Essential for half-angle formulas	Sign ambiguity ($\pm$) leads to incorrect numerical values in exam options.
Polynomial Degree Consistency	`n \ge 1, n \in \mathbb{Z}`	Mandatory for expansion formulas	Failure to account for all roots in Chebyshev-type expansions of $\cos(n\theta)$.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Product of Cosines Series

$$ \cos\theta \cos(2\theta) \cos(4\theta) \cdots \cos(2^{n-1}\theta) = \frac{\sin(2^n\theta)}{2^n \sin\theta} $$

Rule: Multiply and divide by $2\sin\theta$, telescope the numerator using $\sin(2x) = 2\sin x \cos x$.

Application: undefined

Triple Angle Ratios for Special Angles

$$ \sin(3\theta) = 3\sin\theta - 4\sin^3\theta, \quad \cos(3\theta) = 4\cos^3\theta - 3\cos\theta $$

Rule: Use substitution $\theta = 18^\circ$ or $36^\circ$ to construct quadratic/cubic equations for solving numerical values.

Application: Direct evaluation of $\sin 18^\circ = \frac{\sqrt{5}-1}{4}$ and $\cos 36^\circ = \frac{\sqrt{5}+1}{4}$.

Sum of Sine/Cosine with AP Angles

$$ \sum_{k=1}^n \sin(a + k d) = \frac{\sin\left(\frac{nd}{2}\right)}{\sin\left(\frac{d}{2}\right)} \sin\left(a + \frac{(n+1)d}{2}\right) $$

Rule: Apply standard difference formula $\sin\left(\frac{d}{2}\right) [\dots]$ to collapse the telescoping series instantly.

Maximum/Minimum of $a\cos\theta + b\sin\theta$

$$ -\sqrt{a^2 + b^2} \le a\cos\theta + b\sin\theta \le \sqrt{a^2 + b^2} $$

Rule: Instantly read off bounds without calculus or completing squares.

Application: Extremely common in UP TGT/PGT for finding range of trigonometric expressions.

Half-Angle Radical Reduction

$$ \sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos\theta}{2}} $$

Rule: Determine the quadrant of $\frac{\theta}{2}$ first to lock in the correct $\pm$ sign before calculation.

Rapid Elimination via Symmetry

$$ \text{If } \alpha + \beta = \frac{\pi}{2}, \text{ then } \tan\alpha \tan\beta = 1 $$

Rule: Look for complementary angle pairs in product series to eliminate options equal to $0$ or $-1$.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

The Half-Angle Sign Ambiguity

Common Pitfall: Forgetting to check the quadrant of $\frac{\theta}{2}$ when applying half-angle formulas, blindly choosing the positive root.

Counterexample: \text{Evaluating } \sin(100^\circ / 2) = \sin 50^\circ \text{ vs } -\sqrt{\dots}

Why it fails: Leads to incorrect sign if the angle falls in a quadrant where the trigonometric ratio is negative.

🛡️ Exam Defense: Always locate the quadrant of the target sub-multiple angle first.

⚠️ COMMON ERROR #02

Extraneous Roots in Multiple Angle Equations

Common Pitfall: Squaring trigonometric equations or cross-multiplying without checking domain restrictions of denominators.

Counterexample: \tan(2\theta) = \frac{2\tan\theta}{1-\tan^2\theta}

Why it fails: Undefined at $\theta = \frac{\pi}{4} + \frac{n\pi}{2}$, introducing false solutions.

🛡️ Exam Defense: Substitute obtained values back into the original equation or check denominator non-zero condition.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Double Angle	\theta \in \mathbb{R}	\sin 2\theta = 2\sin\theta\cos\theta	Scaling frequency on unit circle	Direct substitution or simplification
Triple Angle	\theta \in \mathbb{R}	\sin 3\theta = 3\sin\theta - 4\sin^3\theta	Cubic polynomial root connection	Problems involving $18^\circ$ or $36^\circ$
Half Angle	\theta/2 \text{ valid quadrant}	\cos\theta = 2\cos^2(\theta/2) - 1	Chord length bisection	Evaluation of radical trigonometric values
Cosine Product Series	n \text{ finite terms}	\prod_{k=0}^{n-1} \cos(2^k\theta) = \frac{\sin(2^n\theta)}{2^n\sin\theta}	Telescoping product collapse	Nested cosine products

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If $\cos\theta = \frac{1}{4}$, find the value of $\cos(3\theta)$.

(A) \frac{11}{16}(B) -\frac{11}{16}(C) \frac{3}{4}(D) -\frac{3}{4}

📚 Method A: Formal Textbook Derivation

Use the triple angle formula: $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$. Substitute $\cos\theta = \frac{1}{4}$: $4\left(\frac{1}{64}\right) - 3\left(\frac{1}{4}\right) = \frac{1}{16} - \frac{3}{4} = \frac{1 - 12}{16} = -\frac{11}{16}$.

⚡ Method B: 25-Second Exam Shortcut

Direct formula plug-in with arithmetic check on fractions.

💡 Tip: Watch out for sign errors when subtracting fractions.

CANONICAL PROBLEM 02 Dual-Method Breakdown

The value of $\cos 20^\circ \cos 40^\circ \cos 80^\circ$ is:

(A) \frac{1}{4}(B) \frac{1}{8}(C) \frac{\sqrt{3}}{8}(D) \frac{1}{16}

📚 Method A: Formal Textbook Derivation

Apply the finite cosine product formula with $\theta = 20^\circ$ and $n=3$: $\cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{\sin(2^3 \times 20^\circ)}{2^3 \sin 20^\circ} = \frac{\sin(160^\circ)}{8 \sin 20^\circ} = \frac{\sin(180^\circ - 20^\circ)}{8 \sin 20^\circ} = \frac{\sin 20^\circ}{8 \sin 20^\circ} = \frac{1}{8}$.

⚡ Method B: 25-Second Exam Shortcut

Recognize the standard pattern $\cos\theta \cos(60^\circ - \theta) \cos(60^\circ + \theta) = \frac{1}{4}\cos(3\theta)$. Here $\cos 20^\circ \cos 40^\circ \cos 80^\circ = \cos 20^\circ \cos(60^\circ-20^\circ) \cos(60^\circ+20^\circ) = \frac{1}{4}\cos(60^\circ) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$.

💡 Tip: The standard product rule is faster than the sine multiplier formula.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If $\tan\theta = \frac{1}{2}$, then $\tan(2\theta)$ is equal to:

A $\frac{3}{4}$

B $\frac{4}{3}$

C $\frac{1}{3}$

D $\frac{3}{2}$

Derivation & Analysis:

Use $\tan(2\theta) = \frac{2\tan\theta}{1-\tan^2\theta} = \frac{2(1/2)}{1 - (1/4)} = \frac{1}{3/4} = \frac{4}{3}$. Correct option is (B).

MCQ 2: The exact value of $\sin 18^\circ$ is:

A $\frac{\sqrt{5}-1}{4}$

B $\frac{\sqrt{5}+1}{4}$

C $\frac{\sqrt{5}-1}{2}$

D $\frac{5-\sqrt{1}}{4}$

Derivation & Analysis:

Standard result derived by setting $\theta = 18^\circ \implies 5\theta = 90^\circ \implies 2\theta = 90^\circ - 3\theta$, yielding $\sin 18^\circ = \frac{\sqrt{5}-1}{4}.

MCQ 3: What is the maximum value of $3\sin\theta + 4\cos\theta$?

A 3

B 4

C 5

D 7

Derivation & Analysis:

Maximum value of $a\cos\theta + b\sin\theta$ is $\sqrt{a^2 + b^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$.

MCQ 4: If $\cos\alpha + \cos\beta = 0$ and $\sin\alpha + \sin\beta = 0$, then $\cos(\alpha - \beta)$ is:

A 0

B 1

C -1

D $\frac{1}{2}$

Derivation & Analysis:

From given conditions, $\cos\alpha = -\cos\beta$ and $\sin\alpha = -\sin\beta$, which implies $\alpha and \beta$ differ by an odd multiple of $\pi$, making $\cos(\alpha - \beta) = \cos(\pi) = -1$ or similar, wait: $\cos(\alpha-\beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta = (-\cos\beta)\cos\beta + (-\sin\beta)\sin\beta = -(\cos^2\beta + \sin^2\beta) = -1$. Wait, let's check correct option index. Option (C) is -1. Let's assign correct index 2.

MCQ 5: The value of $\frac{1 - \cos 2\theta + \sin 2\theta}{1 + \cos 2\theta + \sin 2\theta}$ is equal to:

A $\tan\theta$

B $\cot\theta$

C $\sin\theta$

D $\cos\theta$

Derivation & Analysis:

Substitute $1 - \cos 2\theta = 2\sin^2\theta$, $\sin 2\theta = 2\sin\theta\cos\theta$, $1 + \cos 2\theta = 2\cos^2\theta$. Expression becomes $\frac{2\sin\theta(\sin\theta + \cos\theta)}{2\cos\theta(\cos\theta + \sin\theta)} = \frac{\sin\theta}{\cos\theta} = \tan\theta$.

MCQ 6: If $2 \cos\theta = x + \frac{1}{x}$, then $2 \cos(2\theta)$ equals:

A $x^2 + \frac{1}{x^2}$

B $x + \frac{1}{x}$

C $x^2 - \frac{1}{x^2}$

D 0

Derivation & Analysis:

Using $\cos(2\theta) = 2\cos^2\theta - 1 = 2\left(\frac{x + 1/x}{2}\right)^2 - 1 = \frac{x^2 + 2 + 1/x^2}{2} - 1 = \frac{x^2 + 1/x^2}{2}$, so $2\cos(2\theta) = x^2 + \frac{1}{x^2}.

MCQ 7: The value of $\sin^2 24^\circ - \sin^2 6^\circ$ is:

A $\frac{\sqrt{5}-1}{8}$

B $\frac{\sqrt{5}+1}{8}$

C $\frac{\sqrt{5}-1}{4}$

D $\frac{\sqrt{5}}{8}$

Derivation & Analysis:

Use $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B) = \sin(30^\circ)\sin(18^\circ) = \left(\frac{1}{2}\right) \left(\frac{\sqrt{5}-1}{4}\right) = \frac{\sqrt{5}-1}{8}.

MCQ 8: Assertion (A): $\cos 36^\circ \cos 72^\circ = \frac{1}{4}$. Reason (R): $\cos\theta \cos(60^\circ-\theta)\cos(60^\circ+\theta) = \frac{1}{4}\cos(3\theta)$.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

Both statements are standard identities. Setting $\theta = 36^\circ$ in (R) gives $\cos 36^\circ \cos 24^\circ \cos 96^\circ \dots$ wait, let's verify: $\cos 36^\circ \cos 72^\circ = \frac{\sin 72^\circ \cos 36^\circ \cos 72^\circ}{\sin 72^\circ} = \frac{\sin 144^\circ}{2 \sin 72^\circ} = \frac{1}{2}$? Wait, $\cos 36^\circ = \frac{\sqrt{5}+1}{4}$ and $\cos 72^\circ = \frac{\sqrt{5}-1}{4}$, product is $\frac{5-1}{16} = \frac{4}{16} = \frac{1}{4}$. Thus Assertion is true and Reason correctly explains it.

MCQ 9: If $\tan\left(\frac{\pi}{4} + \theta\right) + \tan\left(\frac{\pi}{4} - \theta\right) = k$, then $k$ equals:

A $2 \sec(2\theta)$

B $2 \tan(2\theta)$

C $2 \cos(2\theta)$

D $2 \csc(2\theta)$

Derivation & Analysis:

Expand using $\tan(A\pm B)$: $\frac{1+\tan\theta}{1-\tan\theta} + \frac{1-\tan\theta}{1+\tan\theta} = \frac{(1+\tan\theta)^2 + (1-\tan\theta)^2}{1-\tan^2\theta} = \frac{2(1+\tan^2\theta)}{1-\tan^2\theta} = \frac{2}{\cos(2\theta)} = 2\sec(2\theta)$.

Q10

MCQ 10: The value of $\sqrt{3} \csc 20^\circ - \sec 20^\circ$ is:

A 1

B 2

C 4

D 0

Derivation & Analysis:

Rewrite as $\frac{\sqrt{3}}{\sin 20^\circ} - \frac{1}{\cos 20^\circ} = \frac{\sqrt{3}\cos 20^\circ - \sin 20^\circ}{\sin 20^\circ \cos 20^\circ} = \frac{2(\frac{\sqrt{3}}{2}\cos 20^\circ - \frac{1}{2}\sin 20^\circ)}{\frac{1}{2}\sin 40^\circ} = \frac{2\sin(60^\circ - 20^\circ)}{\frac{1}{2}\sin 40^\circ} = \frac{2\sin 40^\circ}{\frac{1}{2}\sin 40^\circ} = 4$.

Common questions

Because they bridge quadratic equations and trigonometric ratios without requiring calculators, testing both algebraic manipulation and trigonometric identities in a single step.

Always determine which quadrant $\frac{\theta}{2}$ lies in before applying the square root. For instance, if $\theta = 210^\circ$, then $\frac{\theta}{2} = 105^\circ$ (Quadrant II), where cosine is negative.

Yes, $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$ generates all multiple angle expansions, but memorizing standard results for $2\theta$ and $3\theta$ saves crucial time.