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WBB • Class XI • Mathematics • Ch 3
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Trigonometric Functions

Trigonometry transitions in Class 11 from static right-triangle ratio geometry to dynamic real-valued periodic functions defined on the continuous Cartesian unit circle. Angles are measured in both sexagesimal degrees and circular radians, where one radian represents the angle subtended at the center of a circle by an arc equal in length to its radius, establishing the foundational metric relation s = rθ. On the unit circle x² + y² = 1, any angle θ generates a terminal point P(x, y) = (cos θ, sin θ), extending trigonometric functions to arbitrary positive, negative, and multi-revolution angles. The signs of the six circular functions across the four quadrants are governed by the ASTC rule (All, Sin, Tan, Cos). The core functions exhibit essential harmonic periodicities: sine and cosine repeat with fundamental period 2π and range [-1, 1], whereas tangent repeats with period π across its domain ℝ minus odd multiples of π/2. Analytical trigonometry builds systematically upon compound angle addition theorems, transformation formulas converting products to sums and sums to products (the C-D formulas), and multiple and submultiple angle identities. Finally, trigonometric equations transcend isolated arithmetic solutions to yield infinite solution sets characterized by general solutions with integral parameters n ∈ ℤ.

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Why This Chapter Matters

Trigonometric functions form the universal mathematical language of periodic phenomena, oscillations, wave mechanics, and calculus across mathematics, physics, and electrical engineering. In differential and integral calculus, trigonometric substitutions unlock complex integrals, while the derivatives of sine and cosine drive ordinary differential equations governing harmonic oscillators, alternating currents (AC), electromagnetic radiation, and quantum wavefunctions. In higher secondary competitive examinations such as WBJEE, JEE Main, and JEE Advanced, trigonometric identities and general equation solutions appear in 4 to 8 marks of direct and multi-concept questions linked to limits, coordinate geometry, and complex numbers. In modern technology, the Discrete Cosine Transform (DCT) powers JPEG image compression and MP3 audio encoding, while computer animation engines rely on trigonometric rotations to render 3D scenes in real time. Mastering transformation identities, quadrant reductions, and general solution formulas is indispensable for high academic achievement in the WBCHSE Class 11 annual board examination.

Chapter Roadmap & Progression

1 1. Concept of Angle, Sexagesimal &...
2 2. Unit Circle Definition, Quadrant...
3 3. Trigonometric Functions of Allie...
4 4. Compound Angles (Addition & Subt...
5 5. Transformation Formulas, Multipl...
6 6. Trigonometric Equations & Master...

Complete Concept Guide (100% Curriculum Coverage)

1. Concept of Angle, Sexagesimal & Circular Radian Measures

In elementary geometry, an angle is considered the static divergence between two intersecting rays. In higher mathematics, an angle is generated by rotating a ray about its initial vertex from an initial side to a terminal side.

1.1 Direction & Sign Convention of Angles
  • Positive Angle: Generated when the rotating ray rotates in an anti-clockwise (counter-clockwise) direction.
  • Negative Angle: Generated when the rotating ray rotates in a clockwise direction.
  • Unlike Euclidean geometry where angles are confined between $0^\circ$ and $360^\circ$, trigonometric angles can take any real value from $-\infty$ to $+\infty$ depending on the direction and number of complete rotations.
1.2 Systems of Angular Measurement
System Base Unit Subdivisions & Equivalences
Sexagesimal (English System) Degree ($^\circ$) 1 right angle = $90^\circ$, $1^\circ = 60'$ (minutes), $1' = 60''$ (seconds).
Circular (Radian System) Radian ($\text{rad}$ or $^c$) Constant angle subtended at the center of a circle by an arc of length equal to its radius.
1.3 Theorem: The Radian is a Constant Angle

In any circle of radius $r$, consider an arc $AB$ of length equal to $r$. By definition, the central angle $\angle AOB = 1 \text{ radian}$.

Since angles at the center of a circle are directly proportional to the lengths of the subtending arcs:

$$\frac{\angle AOB}{\text{Straight Angle}} = \frac{\text{Arc } AB}{\text{Semi-circumference}} = \frac{r}{\pi r} = \frac{1}{\pi}$$ $$\implies 1 \text{ radian} = \frac{180^\circ}{\pi} \quad \text{and} \quad \pi \text{ radians} = 180^\circ$$

Numerical approximations:

  • $1 \text{ radian} = \frac{180^\circ}{\pi} \approx 57^\circ 17' 44.8'' \approx 57^\circ 16' 22''$ (using $\pi \approx 22/7$).
  • $1^\circ = \frac{\pi}{180} \text{ radian} \approx 0.017453 \text{ radian}$.
1.4 Arc Length & Sector Area Formulas

Let a circle have radius $r$, and let a central angle $\theta$ be measured strictly in radians:

  1. Arc Length Formula: $$\mathbf{s = r\theta} \iff \theta = \frac{s}{r} \quad (\theta \text{ in radians})$$
  2. Area of Sector: $$\mathbf{\text{Area} = \frac{1}{2} r^2 \theta = \frac{1}{2} r s}$$
Examiner Warning: The formula $s = r\theta$ is invalid if $\theta$ is substituted in degrees! You must always convert degrees to radians first: $\theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180}$.

2. Unit Circle Definition, Quadrants & ASTC Rule

To define trigonometric functions for all real numbers without restriction to acute angles in right triangles, we utilize the Cartesian Unit Circle.

2.1 The Unit Circle Definition of Circular Functions

Consider a circle of radius $r = 1$ centered at the origin $(0, 0)$ in the Cartesian plane, having equation $x^2 + y^2 = 1$. Let an angle $\theta$ have its vertex at the origin and initial side along the positive $x$-axis. Let the terminal side intersect the unit circle at point $P(x, y)$.

Fundamental Coordinate Definitions: $$\mathbf{x = \cos \theta}, \quad \mathbf{y = \sin \theta}$$ $$\tan \theta = \frac{y}{x} = \frac{\sin \theta}{\cos \theta} \quad (x \neq 0)$$ $$\cot \theta = \frac{x}{y} = \frac{\cos \theta}{\sin \theta} \quad (y \neq 0), \quad \sec \theta = \frac{1}{x} = \frac{1}{\cos \theta}, \quad \csc \theta = \frac{1}{y} = \frac{1}{\sin \theta}$$

Since $P(x, y)$ lies on $x^2 + y^2 = 1$, substituting yields the Fundamental Pythagorean Identity:

$$\cos^2 \theta + \sin^2 \theta = 1 \quad \forall \theta \in \mathbb{R}$$
2.2 Signs of Trigonometric Functions Across Quadrants: The ASTC Rule

As the point $P(x, y)$ moves around the four quadrants, the signs of $x$ and $y$ determine the signs of the circular functions:

Quadrant Angle Range Coordinates Positive Functions Mnemonic Rule
Quadrant I $0 < \theta < \pi/2$ $x > 0, y > 0$ ALL ($\sin, \cos, \tan, \cot, \sec, \csc$) All
Quadrant II $\pi/2 < \theta < \pi$ $x < 0, y > 0$ Sine and Cosecant Silver / Sugar
Quadrant III $\pi < \theta < 3\pi/2$ $x < 0, y < 0$ Tangent and Cotangent Tea
Quadrant IV $3\pi/2 < \theta < 2\pi$ $x > 0, y < 0$ Cosine and Secant Cups

Mnemonic: "All Silver Tea Cups" or "Add Sugar To Coffee" (Bengali: অল-সাইন-ট্যান-কস).

2.3 Domain, Range & Periodicity of Trigonometric Functions
Function Natural Domain Range Fundamental Period ($T$)
$f(x) = \sin x$ $\mathbb{R}$ $[-1, 1]$ $2\pi$ ($360^\circ$)
$f(x) = \cos x$ $\mathbb{R}$ $[-1, 1]$ $2\pi$ ($360^\circ$)
$f(x) = \tan x$ $\mathbb{R} - \{(2n+1)\frac{\pi}{2} : n \in \mathbb{Z}\}$ $\mathbb{R}$ ($(-\infty, \infty)$) $\pi$ ($180^\circ$)
$f(x) = \csc x$ $\mathbb{R} - \{n\pi : n \in \mathbb{Z}\}$ $(-\infty, -1] \cup [1, \infty)$ $2\pi$ ($360^\circ$)
$f(x) = \sec x$ $\mathbb{R} - \{(2n+1)\frac{\pi}{2} : n \in \mathbb{Z}\}$ $(-\infty, -1] \cup [1, \infty)$ $2\pi$ ($360^\circ$)
$f(x) = \cot x$ $\mathbb{R} - \{n\pi : n \in \mathbb{Z}\}$ $\mathbb{R}$ ($(-\infty, \infty)$) $\pi$ ($180^\circ$)
2.4 Even and Odd Symmetries
  • Even Functions (Symmetric about $y$-axis): $$\cos(-\theta) = \cos \theta, \quad \sec(-\theta) = \sec \theta$$
  • Odd Functions (Symmetric about origin): $$\sin(-\theta) = -\sin \theta, \quad \tan(-\theta) = -\tan \theta, \quad \csc(-\theta) = -\csc \theta, \quad \cot(-\theta) = -\cot \theta$$

3. Trigonometric Functions of Allied Angles (Reduction Formulas)

Two angles are termed allied angles if their sum or difference is either zero or an integral multiple of a right angle ($\frac{\pi}{2}$ or $90^\circ$).

3.1 The Universal Reduction Algorithm for $n \cdot \frac{\pi}{2} \pm \theta$

To evaluate any trigonometric expression of the form $T\left(n \cdot \frac{\pi}{2} \pm \theta\right)$ where $n \in \mathbb{Z}$ and $\theta$ is an acute angle:

  1. Step 1: Check the Parity of $n$:
    • If $n$ is EVEN ($n = 0, \pm 2, \pm 4, \dots$): The trigonometric function remains unchanged ($\sin \to \sin$, $\cos \to \cos$, $\tan \to \tan$).
    • If $n$ is ODD ($n = \pm 1, \pm 3, \pm 5, \dots$): The trigonometric function transforms into its co-function: $$\sin \longleftrightarrow \cos, \quad \tan \longleftrightarrow \cot, \quad \sec \longleftrightarrow \csc$$
  2. Step 2: Determine the Sign (+ or -): Assume $\theta$ is acute ($0 < \theta < \pi/2$). Locate the quadrant in which the compound angle $\left(n \cdot \frac{\pi}{2} \pm \theta\right)$ terminates. Prefix the sign (+ or -) that the original function possesses in that quadrant according to the ASTC rule.
3.2 Standard Reduction Table Summary
Angle ($\alpha$) $\sin \alpha$ $\cos \alpha$ $\tan \alpha$ Quadrant of $\alpha$
$-\theta$ $-\sin \theta$ $+\cos \theta$ $-\tan \theta$ Quadrant IV
$\frac{\pi}{2} - \theta$ ($90^\circ - \theta$) $+\cos \theta$ $+\sin \theta$ $+\cot \theta$ Quadrant I
$\frac{\pi}{2} + \theta$ ($90^\circ + \theta$) $+\cos \theta$ $-\sin \theta$ $-\cot \theta$ Quadrant II
$\pi - \theta$ ($180^\circ - \theta$) $+\sin \theta$ $-\cos \theta$ $-\tan \theta$ Quadrant II
$\pi + \theta$ ($180^\circ + \theta$) $-\sin \theta$ $-\cos \theta$ $+\tan \theta$ Quadrant III
$\frac{3\pi}{2} - \theta$ ($270^\circ - \theta$) $-\cos \theta$ $-\sin \theta$ $+\cot \theta$ Quadrant III
$\frac{3\pi}{2} + \theta$ ($270^\circ + \theta$) $-\cos \theta$ $+\sin \theta$ $-\cot \theta$ Quadrant IV
$2\pi - \theta$ ($360^\circ - \theta$) $-\sin \theta$ $+\cos \theta$ $-\tan \theta$ Quadrant IV
$2k\pi + \theta$ (Multi-revolution) $+\sin \theta$ $+\cos \theta$ $+\tan \theta$ Quadrant I

4. Compound Angles (Addition & Subtraction Theorems)

An algebraic sum or difference of two or more angles (such as $A + B$ or $A - B$) is called a compound angle.

4.1 Fundamental Addition & Subtraction Theorems
  1. Cosine Theorems: $$\cos(A + B) = \cos A \cos B - \sin A \sin B$$ $$\cos(A - B) = \cos A \cos B + \sin A \sin B$$
  2. Sine Theorems: $$\sin(A + B) = \sin A \cos B + \cos A \sin B$$ $$\sin(A - B) = \sin A \cos B - \cos A \sin B$$
  3. Tangent Theorems: $$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$ $$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$
  4. Cotangent Theorems: $$\cot(A + B) = \frac{\cot A \cot B - 1}{\cot B + \cot A}$$ $$\cot(A - B) = \frac{\cot A \cot B + 1}{\cot B - \cot A}$$
4.2 Special Compound Angle Exact Values

Using $45^\circ - 30^\circ = 15^\circ$ and $45^\circ + 30^\circ = 75^\circ$:

  • $\sin 15^\circ = \sin(45^\circ - 30^\circ) = \frac{\sqrt{6} - \sqrt{2}}{4} = \cos 75^\circ$
  • $\cos 15^\circ = \cos(45^\circ - 30^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4} = \sin 75^\circ$
  • $\tan 15^\circ = \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}} = 2 - \sqrt{3} = \cot 75^\circ$
  • $\tan 75^\circ = 2 + \sqrt{3} = \cot 15^\circ$
4.3 The Product Identity Identity for Sine
$$\sin(A + B)\sin(A - B) = \sin^2 A - \sin^2 B = \cos^2 B - \cos^2 A$$ $$\cos(A + B)\cos(A - B) = \cos^2 A - \sin^2 B = \cos^2 B - \sin^2 A$$

5. Transformation Formulas, Multiple & Submultiple Angles

Calculus integration frequently requires converting products of trigonometric ratios into linear sums, and vice versa.

5.1 Product-to-Sum Transformation Formulas
$$2 \sin A \cos B = \sin(A + B) + \sin(A - B)$$ $$2 \cos A \sin B = \sin(A + B) - \sin(A - B)$$ $$2 \cos A \cos B = \cos(A + B) + \cos(A - B)$$ $$2 \sin A \sin B = \cos(A - B) - \cos(A + B)$$
5.2 Sum-to-Product Formulas (The C-D Formulas)

Let $A + B = C$ and $A - B = D$, so $A = \frac{C + D}{2}$ and $B = \frac{C - D}{2}$:

$$\sin C + \sin D = 2 \sin\frac{C + D}{2} \cos\frac{C - D}{2}$$ $$\sin C - \sin D = 2 \cos\frac{C + D}{2} \sin\frac{C - D}{2}$$ $$\cos C + \cos D = 2 \cos\frac{C + D}{2} \cos\frac{C - D}{2}$$ $$\cos C - \cos D = 2 \sin\frac{C + D}{2} \sin\frac{D - C}{2} = -2 \sin\frac{C + D}{2} \sin\frac{C - D}{2}$$
5.3 Double-Angle Formulas ($2\theta$)
  1. $\sin 2\theta = 2 \sin \theta \cos \theta = \frac{2\tan \theta}{1 + \tan^2 \theta}$
  2. $\cos 2\theta = \cos^2 \theta - \sin^2 \theta = 2\cos^2 \theta - 1 = 1 - 2\sin^2 \theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$
  3. The Power Reduction Lemmas: $$\mathbf{1 - \cos 2\theta = 2 \sin^2 \theta} \iff \sin^2 \theta = \frac{1 - \cos 2\theta}{2}$$ $$\mathbf{1 + \cos 2\theta = 2 \cos^2 \theta} \iff \cos^2 \theta = \frac{1 + \cos 2\theta}{2}$$ $$\frac{1 - \cos 2\theta}{1 + \cos 2\theta} = \tan^2 \theta$$
  4. $\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}$
5.4 Triple-Angle Formulas ($3\theta$)
  • $\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta$
  • $\cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta$
  • $\tan 3\theta = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}$
5.5 Submultiple Angle Values ($18^\circ$ and $36^\circ$)
  • $\sin 18^\circ = \frac{\sqrt{5} - 1}{4} = \cos 72^\circ$
  • $\cos 36^\circ = \frac{\sqrt{5} + 1}{4} = \sin 54^\circ$
  • $\cos 18^\circ = \frac{\sqrt{10 + 2\sqrt{5}}}{4} = \sin 72^\circ$
  • $\sin 36^\circ = \frac{\sqrt{10 - 2\sqrt{5}}}{4} = \cos 54^\circ$

6. Trigonometric Equations & Master General Solutions

An equation containing trigonometric functions of an unknown variable is called a trigonometric equation. Because trigonometric functions are periodic, their equations generally have infinitely many solutions.

6.1 Principal Solution vs. General Solution
  • Principal Solution: Solutions lying in the primary interval $[0, 2\pi)$ or $[0^\circ, 360^\circ)$. A trigonometric equation typically has two principal solutions.
  • General Solution: An algebraic expression involving an arbitrary integer parameter $n \in \mathbb{Z}$ that encapsulates all possible solutions of the equation across $(-\infty, \infty)$.
6.2 Fundamental General Solution Theorems
Equation Form General Solution Formula ($n \in \mathbb{Z}$) Remarks / Derivation Rationale
$\sin \theta = 0$ $\mathbf{\theta = n\pi}$ Zeros at $0, \pm\pi, \pm 2\pi, \dots$
$\cos \theta = 0$ $\mathbf{\theta = (2n + 1)\frac{\pi}{2}}$ Odd multiples of $\pi/2$
$\tan \theta = 0$ $\mathbf{\theta = n\pi}$ Numerator $\sin\theta = 0$
$\sin \theta = \sin \alpha$ $\mathbf{\theta = n\pi + (-1)^n \alpha}$ $(-1)^n$ accommodates Q I / Q II alternating signs
$\cos \theta = \cos \alpha$ $\mathbf{\theta = 2n\pi \pm \alpha}$ Even symmetry $\cos(-\alpha) = \cos \alpha$
$\tan \theta = \tan \alpha$ $\mathbf{\theta = n\pi + \alpha}$ Periodicity of tangent is $\pi$
$\sin^2 \theta = \sin^2 \alpha$
$\cos^2 \theta = \cos^2 \alpha$
$\tan^2 \theta = \tan^2 \alpha$		 $\mathbf{\theta = n\pi \pm \alpha}$ Unified square theorem; reduces via $1 - \cos 2\theta$
6.3 Linear Auxiliary Equation: $a \cos \theta + b \sin \theta = c$

To solve the linear combination $a \cos \theta + b \sin \theta = c$:

  1. Divide the entire equation by $r = \sqrt{a^2 + b^2}$: $$\frac{a}{\sqrt{a^2 + b^2}} \cos \theta + \frac{b}{\sqrt{a^2 + b^2}} \sin \theta = \frac{c}{\sqrt{a^2 + b^2}}$$
  2. Set $\cos \phi = \frac{a}{\sqrt{a^2 + b^2}}$ and $\sin \phi = \frac{b}{\sqrt{a^2 + b^2}}$ where $\tan \phi = \frac{b}{a}$.
  3. The equation collapses into: $$\cos(\theta - \phi) = \frac{c}{\sqrt{a^2 + b^2}}$$
  4. Existence Condition for Real Solutions: $$\left|\frac{c}{\sqrt{a^2 + b^2}}\right| \le 1 \iff \mathbf{|c| \le \sqrt{a^2 + b^2}}$$ The minimum value of $a\cos\theta + b\sin\theta$ is $-\sqrt{a^2+b^2}$ and maximum is $+\sqrt{a^2+b^2}$.

Key Formulas, Identities & Theorems

Arc Length & Radian Formula
s = rθ
Central angle θ must be measured strictly in radians; π radians = 180°.
Fundamental Pythagorean Identities
sin²θ + cos²θ = 1
Valid for all permissible real numbers; derived from x² + y² = 1 on the unit circle.
Compound Angle Addition Theorems
sin(A±B), cos(A±B)
Notice the minus sign in cos(A+B); tan(A±B) = (tan A ± tan B)/(1 ∓ tan A tan B).
Multiple & Half-Angle Identities
1 ± cos 2θ
Fundamental power-reduction identities widely used in calculus integration.
Sum-to-Product (C-D) Transformation
C-D Formulas
Crucial trap: note the order D - C or the minus sign in cos C - cos D.
Master General Solutions
θ = nπ + (-1)ⁿα
For tan θ = tan α, θ = nπ + α; parameter n represents any integer (n ∈ ℤ).

Conceptual Solved Examples & Case Studies

Example 1
A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second? [2 marks]
Step-by-Step Solution:
Solution:
Step 1: Calculate revolutions per second: Number of revolutions in 60 seconds (1 minute) = 360. $$\text{Revolutions in 1 second} = \frac{360}{60} = 6 \text{ revolutions}$$
Step 2: Convert revolutions into radian measure: In one complete revolution, the wheel turns through an angle of $2\pi$ radians (or $360^\circ$). $$\text{Angle turned in 1 second} = 6 \times 2\pi = \mathbf{12\pi \text{ radians}}$$
Conclusion: The wheel turns through $12\pi$ radians (approximately $37.7$ radians) in one second.
Example 2
If \(\cos x = -\frac{3}{5}\) and \(x\) lies in the third quadrant, find the values of all other five trigonometric functions. [3 marks]
Step-by-Step Solution:
Solution:
Step 1: Identify quadrant and ASTC signs: Since $x$ lies in Quadrant III ($\pi < x < \frac{3\pi}{2}$):
  • Tangent ($\tan x$) and Cotangent ($\cot x$) are positive.
  • Sine ($\sin x$), Cosine ($\cos x$), Secant ($\sec x$), and Cosecant ($\csc x$) are negative.

Step 2: Find $\sin x$: Using the fundamental identity $\sin^2 x + \cos^2 x = 1$: $$\sin^2 x = 1 - \cos^2 x = 1 - \left(-\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}$$ Since $\sin x < 0$ in Quadrant III: $$\mathbf{\sin x = -\frac{4}{5}}$$
Step 3: Evaluate the remaining four ratios: $$\mathbf{\sec x = \frac{1}{\cos x} = -\frac{5}{3}}$$ $$\mathbf{\csc x = \frac{1}{\sin x} = -\frac{5}{4}}$$ $$\mathbf{\tan x = \frac{\sin x}{\cos x} = \frac{-4/5}{-3/5} = +\frac{4}{3}}$$ $$\mathbf{\cot x = \frac{1}{\tan x} = +\frac{3}{4}}$$
Example 3
Prove that \(\frac{\cos 11^\circ + \sin 11^\circ}{\cos 11^\circ - \sin 11^\circ} = \tan 56^\circ\). [3 marks]
Step-by-Step Solution:
Solution:
Step 1: Divide numerator and denominator by $\cos 11^\circ$: $$\text{LHS} = \frac{\frac{\cos 11^\circ}{\cos 11^\circ} + \frac{\sin 11^\circ}{\cos 11^\circ}}{\frac{\cos 11^\circ}{\cos 11^\circ} - \frac{\sin 11^\circ}{\cos 11^\circ}} = \frac{1 + \tan 11^\circ}{1 - \tan 11^\circ}$$
Step 2: Express 1 as $\tan 45^\circ$: Since $\tan 45^\circ = 1$: $$\text{LHS} = \frac{\tan 45^\circ + \tan 11^\circ}{1 - \tan 45^\circ \cdot \tan 11^\circ}$$
Step 3: Apply the compound tangent addition formula: Using $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ with $A = 45^\circ, B = 11^\circ$: $$\text{LHS} = \tan(45^\circ + 11^\circ) = \mathbf{\tan 56^\circ} = \text{RHS}$$ Hence proved.
Example 4
Prove the identity: \(\frac{\sin 5x - 2\sin 3x + \sin x}{\cos 5x - \cos x} = \tan x\). [4 marks]
Step-by-Step Solution:
Solution:
Step 1: Group the numerator terms to apply C-D formula: $$\text{Numerator} = (\sin 5x + \sin x) - 2\sin 3x$$ Using $\sin C + \sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2}$: $$\sin 5x + \sin x = 2\sin\left(\frac{5x + x}{2}\right)\cos\left(\frac{5x - x}{2}\right) = 2\sin 3x \cos 2x$$ $$\text{Numerator} = 2\sin 3x \cos 2x - 2\sin 3x = 2\sin 3x (\cos 2x - 1)$$ Using $1 - \cos 2x = 2\sin^2 x \implies \cos 2x - 1 = -2\sin^2 x$: $$\text{Numerator} = 2\sin 3x (-2\sin^2 x) = -4\sin 3x \sin^2 x$$
Step 2: Transform the denominator: Using $\cos C - \cos D = -2\sin\frac{C+D}{2}\sin\frac{C-D}{2}$: $$\text{Denominator} = \cos 5x - \cos x = -2\sin\left(\frac{5x + x}{2}\right)\sin\left(\frac{5x - x}{2}\right) = -2\sin 3x \sin 2x$$ Using $\sin 2x = 2\sin x \cos x$: $$\text{Denominator} = -2\sin 3x (2\sin x \cos x) = -4\sin 3x \sin x \cos x$$
Step 3: Evaluate the fraction: $$\frac{\text{Numerator}}{\text{Denominator}} = \frac{-4\sin 3x \sin^2 x}{-4\sin 3x \sin x \cos x} = \frac{\sin x}{\cos x} = \mathbf{\tan x} = \text{RHS}$$ Hence proved.
Example 5
Prove that: \(\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \frac{1}{16}\). [4 marks]
Step-by-Step Solution:
Solution:
Step 1: Substitute the known value $\cos 60^\circ = \frac{1}{2}$: $$\text{LHS} = \frac{1}{2} \left[ \cos 20^\circ \cos 40^\circ \cos 80^\circ \right]$$
Step 2: Group and multiply/divide by 2 to apply product formula: $$\text{LHS} = \frac{1}{4} \left[ (2\cos 40^\circ \cos 20^\circ) \cos 80^\circ \right]$$ Using $2\cos A \cos B = \cos(A + B) + \cos(A - B)$: $$2\cos 40^\circ \cos 20^\circ = \cos(40^\circ + 20^\circ) + \cos(40^\circ - 20^\circ) = \cos 60^\circ + \cos 20^\circ = \frac{1}{2} + \cos 20^\circ$$
Step 3: Expand with $\cos 80^\circ$: $$\text{LHS} = \frac{1}{4} \left[ \left(\frac{1}{2} + \cos 20^\circ\right) \cos 80^\circ \right] = \frac{1}{8} \cos 80^\circ + \frac{1}{4} \cos 20^\circ \cos 80^\circ$$ Multiply and divide the second term by 2: $$\text{LHS} = \frac{1}{8} \cos 80^\circ + \frac{1}{8} [2\cos 80^\circ \cos 20^\circ]$$ $$2\cos 80^\circ \cos 20^\circ = \cos 100^\circ + \cos 60^\circ = \cos(180^\circ - 80^\circ) + \frac{1}{2} = -\cos 80^\circ + \frac{1}{2}$$
Step 4: Combine the terms: $$\text{LHS} = \frac{1}{8} \cos 80^\circ + \frac{1}{8}\left(-\cos 80^\circ + \frac{1}{2}\right) = \frac{1}{8}\left(\frac{1}{2}\right) = \mathbf{\frac{1}{16}} = \text{RHS}$$ Hence proved.
Example 6
Find the general solution of the trigonometric equation: \(\sqrt{3}\cos\theta + \sin\theta = \sqrt{2}\). [5 marks]
Step-by-Step Solution:
Solution:
Step 1: Identify coefficients and calculate $r = \sqrt{a^2 + b^2}$: Here $a = \sqrt{3}$ and $b = 1$. $$r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$$
Step 2: Divide both sides by 2: $$\frac{\sqrt{3}}{2} \cos\theta + \frac{1}{2} \sin\theta = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$$
Step 3: Convert into a single cosine compound angle: Notice that $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$ and $\sin\frac{\pi}{6} = \frac{1}{2}$. $$\cos\theta \cos\frac{\pi}{6} + \sin\theta \sin\frac{\pi}{6} = \frac{1}{\sqrt{2}}$$ Using $\cos(A - B) = \cos A \cos B + \sin A \sin B$: $$\cos\left(\theta - \frac{\pi}{6}\right) = \frac{1}{\sqrt{2}}$$ Since $\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$: $$\cos\left(\theta - \frac{\pi}{6}\right) = \cos\frac{\pi}{4}$$
Step 4: Apply the master general solution formula for cosine: For $\cos X = \cos \alpha \implies X = 2n\pi \pm \alpha$, where $n \in \mathbb{Z}$: $$\theta - \frac{\pi}{6} = 2n\pi \pm \frac{\pi}{4}$$ $$\mathbf{\theta = 2n\pi + \frac{\pi}{6} \pm \frac{\pi}{4}, \quad n \in \mathbb{Z}}$$
Step 5: Detail the two solution branches:
  • Branch 1 (+): $\theta = 2n\pi + \left(\frac{\pi}{6} + \frac{\pi}{4}\right) = \mathbf{2n\pi + \frac{5\pi}{12}}$
  • Branch 2 (-): $\theta = 2n\pi + \left(\frac{\pi}{6} - \frac{\pi}{4}\right) = \mathbf{2n\pi - \frac{\pi}{12}}$

Common Misconceptions & Examiner Traps

Common Misconception

Substituting degrees directly into arc length s = rθ

Scientific Reality & Correction

Always convert degrees to radians first: θ = 60° × (π/180) = π/3 rad. Then s = 5 × (π/3) = 5π/3 cm.

Common Misconception

Assuming sin(A + B) = sin A + sin B (Linearity Fallacy)

Scientific Reality & Correction

Trigonometric functions are non-linear: sin(30° + 60°) = sin 90° = 1, whereas sin 30° + sin 60° = 1/2 + √3/2 ≈ 1.366 ≠ 1. Always use sin(A + B) = sin A cos B + cos A sin B.

Common Misconception

Forgetting the negative sign in the C-D formula for cos C - cos D

Scientific Reality & Correction

cos C - cos D = -2 sin((C+D)/2) sin((C-D)/2) = 2 sin((C+D)/2) sin((D-C)/2). Notice the reversed term D - C.

Common Misconception

Dividing an equation by a trigonometric term and losing roots

Scientific Reality & Correction

Dividing by sin x eliminates the roots where sin x = 0 (x = nπ). Instead, factor: sin x (2 cos x - 1) = 0 ⟹ sin x = 0 or cos x = 1/2.

Common Misconception

Omitting the (-1)ⁿ factor in the general solution of sin θ = sin α

Scientific Reality & Correction

For sine, when n is even (n = 2k), the terminal side is near 0/2π giving θ = 2kπ + α; when n is odd (n = 2k+1), it is near π giving θ = (2k+1)π - α. Combining both yields θ = nπ + (-1)ⁿ α.

Trigonometric Functions, Harmonic Waveforms & Analytical Equations Roadmap

π Trigonometric Functions & Harmonic Geometry WBCHSE Class 11 Mathematics • Unit Circle, Wave Profiles & General Solutions WBCHSE CLASS 11 1. Unit Circle & The ASTC Rule Circular Definition: P(x, y) = (cos θ, sin θ) x y θ P(x, y) Q I Q II Q III Q IV I: ALL (+) All ratios positive II: SIN (+) sin, csc > 0 III: TAN (+) tan, cot > 0 IV: COS (+) cos, sec > 0 Arc Length: s = rθ | Sector Area: ½r²θ 2. Trigonometric Waveforms & Graphs Periodic Oscillations & Range Profiles y = sin x (Period = 2π) 0 π 2π +1 -1 y = cos x (Period = 2π) 0 π 2π +1 -1 Domain: ℝ | Range: [-1, 1] y = tan x: Range ℝ, Period = π Asymptotes: x = (2n+1)π/2 Even/Odd: cos(-x)=cos x | sin(-x)=-sin x 3. Master Identities & Solutions Compound Angles, Multiples & Equations Compound Angle Addition Laws: sin(A±B) = sin A cos B ± cos A sin B cos(A±B) = cos A cos B ∓ sin A sin B Multiple & Half-Angle Identities: sin 2θ = 2 sin θ cos θ = 2tan θ / (1+tan²θ) 1 - cos 2θ = 2 sin²θ | 1 + cos 2θ = 2 cos²θ General Solutions of Trigonometric Equations: sin θ = sin α ⟹ θ = nπ + (-1)ⁿ α cos θ = cos α ⟹ θ = 2nπ ± α tan θ = tan α ⟹ θ = nπ + α (n ∈ ℤ) Linear auxiliary: a cos θ + b sin θ = c (|c| ≤ √(a²+b²)) Allied Angles Reduction Rule (n·π/2 ± θ): • Even n ⟹ Function remains same • Odd n ⟹ Co-function (sin↔cos, tan↔cot) [ASTC sign]

Chapter Summary & 10 Key Takeaways

Takeaway 1
A radian is the constant angle subtended at the center of a circle by an arc of length equal to its radius; π radians = 180°, and 1 radian ≈ 57° 16' 22''.
Takeaway 2
For a circle of radius r, arc length s subtending central angle θ (in radians) is given by s = rθ, and sector area is A = ½r²θ = ½rs.
Takeaway 3
On the unit circle, P(θ) = (cos θ, sin θ); hence cos²θ + sin²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = csc²θ for all permissible real angles.
Takeaway 4
Signs in the four quadrants follow the ASTC rule: Q I (All positive), Q II (Sine and Csc positive), Q III (Tan and Cot positive), Q IV (Cos and Sec positive).
Takeaway 5
Cosine and secant are even functions [cos(-x) = cos x, sec(-x) = sec x]; sine, tangent, cosecant, and cotangent are odd functions [sin(-x) = -sin x, tan(-x) = -tan x].
Takeaway 6
For allied angles n(π/2) ± θ: if n is even, the function name is preserved; if n is odd, the function changes to its co-function (sin ↔ cos, tan ↔ cot, sec ↔ csc), with the sign determined by the original function in the operand quadrant.
Takeaway 7
Compound angle theorems: cos(A ± B) = cos A cos B ∓ sin A sin B, and sin(A ± B) = sin A cos B ± cos A sin B.
Takeaway 8
Double-angle relations: sin 2θ = 2 sin θ cos θ = 2tan θ / (1 + tan²θ), and cos 2θ = cos²θ - sin²θ = 2cos²θ - 1 = 1 - 2sin²θ = (1 - tan²θ)/(1 + tan²θ).
Takeaway 9
Fundamental power reduction identities: 1 - cos 2θ = 2 sin²θ, and 1 + cos 2θ = 2 cos²θ; crucial for calculus integration.
Takeaway 10
General solutions for equations: sin θ = sin α ⟹ θ = nπ + (-1)ⁿ α; cos θ = cos α ⟹ θ = 2nπ ± α; tan θ = tan α ⟹ θ = nπ + α, where n ∈ ℤ.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
Find the degree measure corresponding to -4 radians (take π = 22/7).
Reveal Answer & Explanation
Answer: -4 radians = -4 × (180/π)° = -4 × (180 × 7 / 22)° = -2520/11° = -229° + (1/11)° = -229° 5' 27''.
Multiply by 180/π and convert fractional degrees into minutes (' ) and seconds ('' ) by multiplying by 60 successively.
2
Find the value of sin(31π/3).
Reveal Answer & Explanation
Answer: sin(31π/3) = sin(10π + π/3) = sin(5 × 2π + π/3) = sin(π/3) = √3/2.
Write 31π/3 as an even multiple of π plus an acute angle: 10π + π/3.
3
Find the maximum and minimum values of the expression 3 sin x + 4 cos x + 7.
Reveal Answer & Explanation
Answer: For 3 sin x + 4 cos x, maximum is √(3² + 4²) = 5, and minimum is -5. Therefore, Maximum = 5 + 7 = 12, and Minimum = -5 + 7 = 2.
The range of a sin x + b cos x is [-√(a²+b²), +√(a²+b²)]. Add 7 to both endpoints.
4
If tan A = 1/2 and tan B = 1/3, find the value of A + B (where A, B are positive acute angles).
Reveal Answer & Explanation
Answer: tan(A + B) = (tan A + tan B) / (1 - tan A tan B) = (1/2 + 1/3) / (1 - 1/2 × 1/3) = (5/6) / (5/6) = 1. Since A, B are acute, A + B = π/4 (or 45°).
Use the compound angle formula tan(A + B) = (tan A + tan B) / (1 - tan A tan B).
5
Find the general solution of the equation tan 2x = -cot(x + π/3).
Reveal Answer & Explanation
Answer: tan 2x = -cot(x + π/3) = tan(π/2 + x + π/3) = tan(x + 5π/6). Using tan θ = tan α ⟹ θ = nπ + α: 2x = nπ + x + 5π/6 ⟹ x = nπ + 5π/6, where n ∈ ℤ.
Convert -cot(A) into tan(π/2 + A) using allied angle reduction.
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