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10. Trigonometry & Complex Numbers

Trigonometric equations

Trigonometric equations involve unknown angles constrained by trigonometric functions, possessing infinite periodic solutions governed by fundamental principal values. This topic covers general solutions and interval-restricted root counting.

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

General Solutions of Standard Trigonometric Equations
For standard trigonometric equations, the general solutions are given by: (1) $\sin \theta = \sin \alpha \implies \theta = n\pi + (-1)^n \alpha$, (2) $\cos \theta = \cos \alpha \implies \theta = 2n\pi \pm \alpha$, (3) $\tan \theta = \tan \alpha \implies \theta = n\pi + \alpha$, where $n \in \mathbb{Z}$ and $\alpha \in [-\pi/2, \pi/2]$ or corresponding principal domains.

📐 Geometric Interpretation

Intersections of periodic sinusoidal curves with horizontal constant lines $y = k$ within their range $[-1, 1]$, exhibiting infinite symmetrical recurrence across the real number line $\mathbb{R}$.

⚙️ Algebraic & Structural Meaning

Establishes a bridge between transcendental trigonometric functions and discrete integer lattice points via periodic mappings in abstract algebra.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Domain Restrictions for Tangent / Secant \theta \neq (2n+1)\frac{\pi}{2} Strictly Mandatory Division by zero or undefined function values (vertical asymptotes).
Range Bounds for Sine and Cosine -1 \le k \le 1 Strictly Mandatory Complex imaginary solutions instead of real solutions if bounds are exceeded.
Principal Value Domain Interval [-\frac{\pi}{2}, \frac{\pi}{2}] \text{ or } [0, \pi] Essential for uniqueness Ambiguity in defining inverse trigonometric values and principal angles $\alpha$.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Auxiliary Angle Transformation for $a \cos \theta + b \sin \theta = c$

$$a \cos \theta + b \sin \theta = c \implies \text{Real solutions exist iff } c^2 \le a^2 + b^2$$

Rule: Check condition $c^2 \le a^2 + b^2$ instantly before attempting factorization or squaring.

Application: Frequently asked in UP PGT to find the number of solutions or maximum/minimum values.
02

Sum of Roots in Given Intervals

$$ \text{For } a \cos \theta + b \sin \theta = c, \text{ set } t = \tan(\theta/2) \implies (c-a)t^2 - 2bt + (c+a) = 0 $$

Rule: Use the Weierstrass substitution $t = \tan(\theta/2)$ to convert trigonometric equations into rational quadratic equations in $t$.

Application: Standard pattern in DSSSB PGT for finding sum of all solutions in $[0, 2\pi]$.
03

Factorization of Symmetric Products

$$ \sin 3\theta = 4 \sin \theta \sin(\frac{\pi}{3} - \theta)\sin(\frac{\pi}{3} + \theta) $$

Rule: Instantly expand product forms into multiple-angle formulas to cancel common terms.

Application: Directly tested in KVS PGT and UP TGT simplification problems.
04

Number of Solutions in $[0, \pi]$ for Multiple Angles

$$ \sin(n\theta) = \sin(m\theta) \text{ or similar forms analyzed via graph intersections} $$

Rule: Count intersections by sketching rough sine/cosine waves inside the bounded interval.

Application: Directly asked in UP PGT 2021 regarding number of roots.
05

Squaring Trap Reversal

$$ \sin \theta + \cos \theta = \sqrt{2}\cos(\theta - \frac{\pi}{4}) $$

Rule: Avoid squaring equations directly to prevent introduction of extraneous roots; use auxiliary angle form instead.

Application: Crucial for eliminating incorrect options in multiple-choice questions.
06

Rapid Elimination via Boundary Substitution

$$ \text{Test } \theta = 0, \frac{\pi}{2}, \pi \text{ to eliminate incorrect options instantly.} $$

Rule: Substitute standard test angles into options to rule out at least two distractors.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Extraneous Roots from Squaring

Common Pitfall: Squaring equations like $\sin \theta + \cos \theta = 1$ introduces extra solutions that do not satisfy the original unsquared equation.

Counterexample: \text{Squaring } \sin \theta - \cos \theta = 1 \implies 1 - \sin 2\theta = 1 \implies \sin 2\theta = 0

Why it fails: Squaring alters the domain of validity, turning negative signs into positive ones.

🛡️ Exam Defense: Always plug candidate roots back into the original equation before finalizing.
⚠️ COMMON ERROR #02

Division by Trigonometric Expressions

Common Pitfall: Dividing both sides by $\cos \theta$ or $\sin \theta$ without checking if the expression equals zero.

Counterexample: \sin \theta = \cos \theta \implies \text{ dividing by } \cos \theta \text{ gives } \tan \theta = 1

Why it fails: If $\cos \theta = 0$, division by zero is illegal and root $\theta = \pi/2 + n\pi$ is permanently lost.

🛡️ Exam Defense: Factor out common terms instead of dividing: $a \sin \theta = b \cos \theta \implies factor out$, never cancel variables.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Sine General Solutionk \in [-1, 1]\theta = n\pi + (-1)^n \alphaHorizontal line cutting sine wave infinitelyEquation involving \sin \theta = \sin \alpha
Cosine General Solutionk \in [-1, 1]\theta = 2n\pi \pm \alphaHorizontal line cutting cosine wave infinitelyEquation involving \cos \theta = \cos \alpha
Tangent General Solution\theta \neq \frac{\pi}{2} + n\pi\theta = n\pi + \alphaIntersections with periodic tan branchesEquation involving \tan \theta = \tan \alpha
Auxiliary Form Bounda^2 + b^2 \ge c^2Real solutions conditionScaling and phase shift of sine waveExpression $a \cos \theta + b \sin \theta = c$

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the general solution of the trigonometric equation $\tan \theta + \tan 2\theta + \tan 3\theta = \tan \theta \tan 2\theta \tan 3\theta$.

(A) \theta = \frac{n\pi}{6}, n \in \mathbb{Z}(B) \theta = \frac{n\pi}{2}, n \in \mathbb{Z}(C) \theta = n\pi, n \in \mathbb{Z}(D) \theta = \frac{n\pi}{3}, n \in \mathbb{Z}
✓ Correct Answer: (D)
📚 Method A: Formal Textbook Derivation
Step 1: Recognize the identity $\tan(A+B+C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - \tan A \tan B - \tan B \tan C - \tan C \tan A}$. Step 2: The given equation implies $\tan(\theta + 2\theta + 3\theta) = \tan(6\theta)$ is undefined or numerator is zero, leading to $\tan 6\theta = 0$. Step 3: $6\theta = n\pi \implies \theta = n\pi / 6$. However, check domain restrictions for individual tangents to exclude extraneous points.
⚡ Method B: 25-Second Exam Shortcut
Note that $\theta + 2\theta + 3\theta = 6\theta$, so the sum of angles corresponds directly to $\tan 6\theta = 0 \implies \theta = \frac{n\pi}{6}$. Refine by checking denominator non-zero.
💡 Tip: Watch out for domain traps where individual tangents become infinity.
CANONICAL PROBLEM 02 Dual-Method Breakdown

If $0 \le \theta \le 2\pi$ and $2 \sin^2 \theta + 3 \sin \theta - 2 = 0$, then the number of solutions is:

(A) 1(B) 2(C) 3(D) 4
✓ Correct Answer: (B)
📚 Method A: Formal Textbook Derivation
Step 1: Let $x = \sin \theta$. The equation becomes quadratic: $2x^2 + 3x - 2 = 0$. Step 2: Factorize: $(2x - 1)(x + 2) = 0 \implies x = 1/2$ or $x = -2$. Step 3: Since $-1 \le \sin \theta \le 1$, $x = -2$ is rejected. Step 4: $\sin \theta = 1/2$. In $[0, 2\pi]$, $\theta = \pi/6, 5\pi/6$. Total solutions = 2.
⚡ Method B: 25-Second Exam Shortcut
Instantly discard $x = -2$ using range bounds of sine, then solve $\sin \theta = 1/2$ in first two quadrants.
💡 Tip: Always test quadratic roots against the strict range $[-1, 1]$ for sine and cosine.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: The general solution of $\sin^2 \theta = \sin^2 \alpha$ is given by:
A $\theta = n\pi \pm \alpha$
B $\theta = 2n\pi \pm \alpha$
C $\theta = n\pi + (-1)^n \alpha$
D $\theta = \frac{n\pi}{2} \pm \alpha$
Derivation & Analysis:

Using trigonometric identities, $\sin^2 \theta = \sin^2 \alpha \implies \cos 2\theta = \cos 2\alpha \implies 2\theta = 2n\pi \pm 2\alpha \implies \theta = n\pi \pm \alpha$.

Q2
MCQ 2: The number of real solutions of $\cos(\text{sin } x) = \sin(\text{cos } x)$ is:
A 0
B 1
C 2
D Infinite
Derivation & Analysis:

Since $|\sin x| \le 1$ and $|\cos x| \le 1$, we examine $\cos(\sin x) = \sin(\cos x) = \cos(\frac{\pi}{2} - \cos x)$. This requires $\sin x = \pm \frac{\pi}{2} - \cos x$, which exceeds the bounds of real values since $\frac{\pi}{2} \approx 1.57 > 1$.

Q3
MCQ 3: If $\sec \theta - \tan \theta = \frac{1}{2}$, then the value of $\sec \theta$ is:
A $\frac{5}{4}$
B $\frac{3}{4}$
C $\frac{5}{2}$
D $\frac{3}{2}$
Derivation & Analysis:

We know $\sec \theta + \tan \theta = \frac{1}{\sec \theta - \tan \theta} = 2$. Adding this to $\sec \theta - \tan \theta = \frac{1}{2}$ gives $2 \sec \theta = 2 + \frac{1}{2} = \frac{5}{2} \implies \sec \theta = \frac{5}{4}$.

Q4
MCQ 4: The general solution of $\sqrt{3} \cos \theta + \sin \theta = \sqrt{2}$ is:
A $\theta = 2n\pi \pm \frac{\pi}{4} - \frac{\pi}{3}$
B $\theta = n\pi + (-1)^n \frac{\pi}{4} + \frac{\pi}{3}$
C $\theta = 2n\pi \pm \frac{\pi}{12}$
D $\theta = 2n\pi \pm \frac{\pi}{4} + \frac{\pi}{3}$
Derivation & Analysis:

Dividing by 2 gives $\frac{\sqrt{3}}{2}\cos \theta + \frac{1}{2}\sin \theta = \frac{\sqrt{2}}{2} \implies \cos(\theta - \pi/3) = \cos(\pi/4) \implies \theta - \pi/3 = 2n\pi \pm \pi/4$.

Q5
MCQ 5: The number of solutions of the equation $\tan \theta + \sec \theta = 2 \cos \theta$ in the interval $[0, 2\pi]$ is:
A 1
B 2
C 3
D 4
Derivation & Analysis:

Convert to sine and cosine: $\frac{\sin \theta + 1}{\cos \theta} = 2 \cos \theta \implies \sin \theta + 1 = 2 \cos^2 \theta = 2(1 - \sin^2 \theta)$. Let $t = \sin \theta \implies 2t^2 + t - 1 = 0 \implies (2t-1)(t+1) = 0 \implies t = 1/2$ or $t = -1$. Checking domains yields 2 valid solutions in $[0, 2\pi]$.

Q6
MCQ 6: If $3 \cos \theta + 4 \sin \theta = \lambda$, for what values of $\lambda$ does the equation have real solutions?
A $-\lambda \le 5$
B $-5 \le \lambda \le 5$
C $\lambda \ge 5$
D All real numbers
Derivation & Analysis:

The maximum and minimum values of $a \cos \theta + b \sin \theta$ are given by $\pm \sqrt{a^2 + b^2} = \pm \sqrt{3^2 + 4^2} = \pm 5$. Thus $-5 \le \lambda \le 5$.

Q7
MCQ 7: The complete set of values of $\theta$ satisfying $\sin \theta \tan \theta = \tan \theta - \sin \theta$ is:
A $\theta = n\pi$
B $\theta = 2n\pi \pm \frac{\pi}{3}$
C $\theta = n\pi \text{ or } \theta = 2n\pi \pm \frac{\pi}{3}$
D $\theta = n\pi + \frac{\pi}{4}$
Derivation & Analysis:

Rewrite as $\sin \theta \tan \theta - \tan \theta + \sin \theta = 0 \implies \tan \theta(\sin \theta - 1) + 1(\sin \theta - 1) = 0$ is incorrect factorization. Proper form: $\sin \theta \frac{\sin \theta}{\cos \theta} = \frac{\sin \theta}{\cos \theta} - \sin \theta \implies \sin^2 \theta = \sin \theta - \sin \theta \cos \theta \implies \sin \theta(\sin \theta - 1 + \cos \theta) = 0$. Solutions are $\sin \theta = 0$ (yielding $\theta = n\pi$) and $\sin \theta + \cos \theta = 1$ (yielding $\theta = 2n\pi$ or $2n\pi + \pi/2$, and $2n\pi \pm \pi/3$ after full reduction).

Q8
MCQ 8: Assertion (A): The equation $\sin \theta = x + \frac{1}{x}$ has no real solution for $x \in \mathbb{R}$. Reason (R): The range of $\sin \theta$ is $[-1, 1]$ while $x + \frac{1}{x} \ge 2$ or $\le -2$ for real $x$.
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:

By AM-GM inequality, $x + 1/x \ge 2$ for $x > 0$ and $\le -2$ for $x < 0$. Thus, $x + 1/x$ can never lie in $[-1, 1]$ except at boundaries if properly constrained, but here no intersection with sine's range $[-1, 1]$ exists except at $\pm 2$, which gives $\sin \theta = \pm 2$, impossible for real $\theta$.

Q9
MCQ 9: If $\cot \theta - \tan \theta = 2$, then the general value of $\theta$ is:
A $\frac{n\pi}{2} + \frac{\pi}{8}$
B $\frac{n\pi}{2} - \frac{\pi}{8}$
C $n\pi \pm \frac{\pi}{4}$
D $\frac{n\pi}{4}$
Derivation & Analysis:

Rewrite as $\frac{\cos \theta}{\sin \theta} - \frac{\sin \theta}{\cos \theta} = 2 \implies \frac{\cos^2 \theta - \sin^2 \theta}{\sin \theta \cos \theta} = 2 \implies \frac{\cos 2\theta}{\frac{1}{2} \sin 2\theta} = 2 \implies \cot 2\theta = 1 \implies 2\theta = n\pi + \frac{\pi}{4} \implies \theta = \frac{n\pi}{2} + \frac{\pi}{8}.

Q10
MCQ 10: The number of roots of $\sin 2\theta - \cos 2\theta = 0$ in the interval $[0, 2\pi]$ is:
A 2
B 4
C 6
D 8
Derivation & Analysis:

$\sin 2\theta = \cos 2\theta \implies \tan 2\theta = 1 \implies 2\theta = n\pi + \frac{\pi}{4} \implies \theta = \frac{n\pi}{2} + \frac{\pi}{8}$. For $n = 0, 1, 2, 3$, we get 4 distinct values inside $[0, 2\pi]$. After checking $n=4$, we exceed $2\pi$. Thus exactly 4 solutions.

Common questions

Extraneous roots typically arise from squaring both sides of an equation or multiplying by expressions containing trigonometric variables that can equal zero. Always verify solutions by substituting them back into the original equation.

Dividing both sides of an equation by a trigonometric term like $\cos \theta$ without checking if $\cos \theta = 0$, which leads to the permanent loss of valid general solutions.

Utilize auxiliary angle transformations ($a \cos \theta + b \sin \theta = \sqrt{a^2+b^2} \cos(\theta - \alpha)$) and test boundary angle options ($0, \pi/2, \pi$) to rapidly eliminate incorrect choices.

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