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अध्याय 3: त्रिकोणमितीय फलन

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CBSE • कक्षा XI • Mathematics • अध्याय 3

अनुमानित समय: 45 Mins

प्रगति: अध्ययनरत

त्रिकोणमितीय फलन

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

अध्याय CBT ऑनलाइन टेस्ट दें सभी अध्याय देखें

प्रस्तावना अध्याय अवधारणाएँ अध्याय सारांश स्व-मूल्यांकन अभ्यास टेस्ट दें

📐 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.

यह अध्याय क्यों महत्वपूर्ण है

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

अध्ययन से पूर्व (आवश्यक ज्ञान)

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

इस अध्याय के लक्ष्य

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.

अध्याय रूपरेखा एवं प्रगति

1 1. Radian Measure & Circular Trigon...

2 2. The ASTC Rule (All Silver Tea Cu...

3 3. Core Trigonometric Identities

सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन

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: $$\mathbf{\cos 2x = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x = \frac{1-\tan^2 x}{1+\tan^2 x}}$$ $$\mathbf{\sin 2x = 2\sin x\cos x = \frac{2\tan x}{1+\tan^2 x}} \quad \text{and} \quad \mathbf{\tan 2x = \frac{2\tan x}{1-\tan^2 x}}$$

चित्रात्मक व्याख्या एवं मॉडल

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)

अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष

मुख्य बिंदु 1

Radian Measure: Natural circular arc-to-radius angular metric where $180^\circ = \pi\text{ rad}$.

मुख्य बिंदु 2

ASTC Rule: Sign convention mnemonic across Quadrants I, II, III, and IV.

मुख्य बिंदु 3

Unit Circle: Geometric foundation equating coordinates $(x, y)$ to $(\cos\theta, \sin\theta)$.

मुख्य बिंदु 4

Double Angle Formulas: Tools reducing powers of trigonometric functions in calculus integration.

मुख्य बिंदु 5

Sum-to-Product: Identities translating product signals into wave sums.

स्व-मूल्यांकन अभ्यास (Check Your Understanding)

मूल वैचारिक स्पष्टता की जांच के लिए नैदानिक प्रश्न। पहले स्वयं हल करें, फिर उत्तर देखें।

Convert $40^\circ 20'$ into radian measure.

उत्तर एवं व्याख्या देखें

उत्तर: $20' = \frac{20}{60}^\circ = \frac{1}{3}^\circ$. 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.

Find the value of $\sin 75^\circ$.

उत्तर एवं व्याख्या देखें

उत्तर: $\sin(45^\circ + 30^\circ) = \sin 45^\circ\cos 30^\circ + \cos 45^\circ\sin 30^\circ = \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}$.
(√3 + 1) / (2√2).

If $\cos x = -\frac{3}{5}$ and $x$ lies in the third quadrant, find the values of other five trigonometric functions.

उत्तर एवं व्याख्या देखें

उत्तर: 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.

Prove that $\cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x}$.

उत्तर एवं व्याख्या देखें

उत्तर: 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.

Find the principal solution of $\tan x = \sqrt{3}$.

उत्तर एवं व्याख्या देखें

उत्तर: $\tan x > 0$ 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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