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CBSE • Class XI • Mathematics • Ch 3
Estimated Time: 45 Mins
Study Progress: In Progress

Trigonometric Functions

In Class 11 Mathematics, "Trigonometric Functions" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

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📐 Have You Ever Wondered?

How do GPS satellites, radio sound waves, and rotating quantum fields oscillate endlessly in rhythmic harmony? Trigonometric functions extended from r...

How do GPS satellites, radio sound waves, and rotating quantum fields oscillate endlessly in rhythmic harmony? Trigonometric functions extended from right triangles to circular radian angles power harmonic analysis.

Why This Chapter Matters

In Class 11 Mathematics, "Trigonometric Functions" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

Before You Begin (Prerequisites)

  • Trigonometric ratios from Class 10.
  • Pythagorean identity.
  • Circle geometry.

What You Will Learn (Core Objectives)

  • Convert between Degree and Radian measure: $\pi\text{ radians} = 180^\circ$ and arc length $l = r\theta$.
  • Define trigonometric functions for arbitrary angles using unit circle coordinates.
  • Determine signs and ranges of trigonometric functions across the four quadrants (ASTC rule).
  • Apply compound angle formulas: $\cos(x \pm y)$, $\sin(x \pm y)$, $\tan(x \pm y)$.
  • Apply double-angle ($2x$), triple-angle ($3x$), and sum-to-product transformation formulas.

Chapter Roadmap & Progression

1 1. Radian Measure & Circular Trigon...
2 2. The ASTC Rule (All Silver Tea Cu...
3 3. Core Trigonometric Identities

Complete Concept Guide (100% Curriculum Coverage)

1. Radian Measure & Circular Trigonometry

An angle subtended at the center of a circle of radius $r$ by an arc of length $l$ is: $$\mathbf{\theta = \frac{l}{r}\text{ radians}} \quad (180^\circ = \pi\text{ rad} \approx 3.14159\text{ rad})$$ On a unit circle ($r=1$), coordinates of any point are $(\cos\theta, \sin\theta)$.

2. The ASTC Rule (All Silver Tea Cups)

Signs across the four quadrants:
• Quadrant I ($0$ to $\pi/2$): All functions are positive.
• Quadrant II ($\pi/2$ to $\pi$): Sin and cosec are positive.
• Quadrant III ($\pi$ to $3\pi/2$): Tan and cot are positive.
• Quadrant IV ($3\pi/2$ to $2\pi$): Cos and sec are positive.

3. Core Trigonometric Identities

  • Compound Angle: $\cos(x+y) = \cos x\cos y - \sin x\sin y$; $\sin(x+y) = \sin x\cos y + \cos x\sin y$.
  • Double Angle: $$\mathbf{\cos 2x = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x = \frac{1-\tan^2 x}{1+\tan^2 x}}$$ $$\mathbf{\sin 2x = 2\sin x\cos x = \frac{2\tan x}{1+\tan^2 x}} \quad \text{and} \quad \mathbf{\tan 2x = \frac{2\tan x}{1-\tan^2 x}}$$

Visual Learning & Conceptual Map

Trigonometric Functions Master Matrix

Conceptual framework, core mechanisms, and analytical relationships
Academic Architecture

1. Radian Measure & Circular Trigonometry • 2. The ASTC Rule (All Silver Tea Cups)

Chapter Summary & 10 Key Takeaways

Takeaway 1
Radian Measure: Natural circular arc-to-radius angular metric where $180^\circ = \pi\text{ rad}$.
Takeaway 2
ASTC Rule: Sign convention mnemonic across Quadrants I, II, III, and IV.
Takeaway 3
Unit Circle: Geometric foundation equating coordinates $(x, y)$ to $(\cos\theta, \sin\theta)$.
Takeaway 4
Double Angle Formulas: Tools reducing powers of trigonometric functions in calculus integration.
Takeaway 5
Sum-to-Product: Identities translating product signals into wave sums.

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
Convert $40^\circ 20'$ into radian measure.
Reveal Answer & Explanation
Answer: $20' = \frac{20}{60}^\circ = \frac{1}{3}^\circ$. So $40^\circ 20' = \frac{121}{3}^\circ$. In radians: $\frac{121}{3} \times \frac{\pi}{180} = \frac{121\pi}{540}\text{ radians}$.
121π / 540 radians.
2
Find the value of $\sin 75^\circ$.
Reveal Answer & Explanation
Answer: $\sin(45^\circ + 30^\circ) = \sin 45^\circ\cos 30^\circ + \cos 45^\circ\sin 30^\circ = \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}$.
(√3 + 1) / (2√2).
3
If $\cos x = -\frac{3}{5}$ and $x$ lies in the third quadrant, find the values of other five trigonometric functions.
Reveal Answer & Explanation
Answer: In Quadrant III, $\tan$ and $\cot$ are positive; others negative. $\sin x = -\sqrt{1 - (-3/5)^2} = -\frac{4}{5}$. $\tan x = \frac{-4/5}{-3/5} = \frac{4}{3}$, $\cot x = \frac{3}{4}$, $\sec x = -\frac{5}{3}$, $\text{cosec } x = -\frac{5}{4}$.
sin=-4/5, tan=4/3, cot=3/4, sec=-5/3, cosec=-5/4.
4
Prove that $\cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x}$.
Reveal Answer & Explanation
Answer: RHS: $\frac{1 - \sin^2 x/\cos^2 x}{1 + \sin^2 x/\cos^2 x} = \frac{\cos^2 x - \sin^2 x}{\cos^2 x + \sin^2 x} = \frac{\cos 2x}{1} = \cos 2x = \text{LHS}$.
Proved using sin/cos expansion.
5
Find the principal solution of $\tan x = \sqrt{3}$.
Reveal Answer & Explanation
Answer: $\tan x > 0$ in Quadrants I and III. In Quadrant I: $x = \frac{\pi}{3}$. In Quadrant III: $x = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$. Principal solutions are $\frac{\pi}{3}$ and $\frac{4\pi}{3}$.
x = π/3 and 4π/3.
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