Simple Learning HubCLASS 9–12 · JEE · NEET

Ch 3Trigonometric Functionsत्रिकोणमितीय फलनTrigonometric Functions

Degree and radian measure, trigonometric functions of any angle on the unit circle, signs in quadrants, graphs, and the sum, difference, double-angle and sum-to-product formulas — with all NCERT exercises solved.अंश और रेडियन माप, इकाई वृत्त पर किसी भी कोण के त्रिकोणमितीय फलन, चतुर्थांशों में चिह्न, आलेख, तथा योग, अंतर, द्विगुण कोण और योग-गुणन सूत्र — NCERT के सभी प्रश्नों के हल।Degree aur radian measure, unit circle par kisi bhi angle ke trigonometric functions, quadrants mein signs, graphs, aur sum, difference, double-angle aur sum-to-product formulas — NCERT ke saare questions ke hal.

Didn’t understand a paragraph? Tap it (or its अ button) → हिंदी / Hinglish / Simple English.कोई पैराग्राफ़ समझ नहीं आया? उस पर (या अ बटन पर) टैप करें → English / Hinglish।Koi paragraph samajh nahi aaya? Us par (ya अ button par) tap kijiye → हिंदी / English.

👆 Words with a open a diagram or graph — tap any label to read what it means.👆 वाले शब्द पर टैप करें — चित्र या ग्राफ़ खुलेगा (हर लेबल पर टैप = उसका अर्थ)।👆 wale word par tap kijiye — diagram ya graph khulega (har label par tap = uska matlab).
📘 NCERT chapter (English PDF)📗 NCERT हिंदी PDF∑ Formulas✎ NCERT solutions🎯 Quiz

Diagrams in this chapterइस अध्याय के चित्रIs chapter ke diagrams

One radian: the arc equal to the radiusएक रेडियन: त्रिज्या के बराबर चापEk radian: radius ke barabar arcFig 3.4Unit circle: cos x and sin x, and signs in the four quadrantsइकाई वृत्त: cos x और sin x, तथा चारों चतुर्थांशों में चिह्नUnit circle: cos x aur sin x, aur chaaron quadrants mein signsFig 3.6Graphs of y = sin x and y = cos xy = sin x और y = cos x के आलेखy = sin x aur y = cos x ke graphsSection 3.3
In this chapter (NCERT sections)इस अध्याय में (NCERT खंड)Is chapter mein (NCERT sections)
  1. 3.2AnglesकोणAngles
  2. 3.3Trigonometric Functionsत्रिकोणमितीय फलनTrigonometric Functions
  3. 3.4Trigonometric Functions of Sum and Difference of Two Anglesदो कोणों के योग और अंतर के त्रिकोणमितीय फलनDo angles ke sum aur difference ke trigonometric functions
  4. ★Yaad rakho — key formulasयाद रखो — मुख्य सूत्रYaad rakho — key formulas
  5. ⚠Common mistakesआम गलतियाँCommon mistakes
  6. ✎NCERT exercises — solutionsNCERT प्रश्नावली — हलNCERT exercises — solutions
  7. 🎯Quizक्विज़Quiz

3.2AnglesकोणAngles

An angle is the amount of rotation of a ray from its initial side to its terminal side. Anticlockwise rotation gives a positive angle, clockwise a negative angle. One complete revolution = 360°.कोण किरण के प्रारंभिक भुजा से अंतिम भुजा तक घूर्णन की माप है। वामावर्त घूर्णन से धनात्मक और दक्षिणावर्त से ऋणात्मक कोण बनता है। एक पूर्ण चक्कर = 360°।Angle ek ray ke initial side se terminal side tak ghoomne ka maap hai. Anticlockwise ghoomne se positive aur clockwise se negative angle banta hai. Ek poora chakkar = 360°.

Degree: 1^\circ = 60' (minutes), 1' = 60'' (seconds). Radian: the angle at the centre made by an arc whose length equals the radius — see the radian picture. For an arc of length l in a circle of radius r: l = r\theta (θ in radians).अंश: 1^\circ = 60' (मिनट), 1' = 60'' (सेकंड)। रेडियन: त्रिज्या के बराबर लंबाई का चाप केंद्र पर जो कोण बनाता है — रेडियन चित्र देखिए। त्रिज्या r के वृत्त में लंबाई l के चाप के लिए: l = r\theta (θ रेडियन में)।Degree: 1^\circ = 60' (minutes), 1' = 60'' (seconds). Radian: radius ke barabar lambai ka arc centre par jo angle banata hai — radian picture dekhiye. Radius r ke circle mein lambai l ke arc ke liye: l = r\theta (θ radian mein).

One radian: the arc equal to the radiusएक रेडियन: त्रिज्या के बराबर चापEk radian: radius ke barabar arcFig 3.4 · 4 tappable partsचित्र 3.4 · 4 टैप करने लायक हिस्सेFig 3.4 · 4 tappable partsOpen
\pi \text{ radian} = 180^\circ\pi \text{ रेडियन} = 180^\circ\pi \text{ radian} = 180^\circ
The key conversionमूल रूपांतरणMain conversion
\text{Radian} = \frac{\pi}{180} \times \text{Degree}\text{रेडियन} = \frac{\pi}{180} \times \text{अंश}\text{Radian} = \frac{\pi}{180} \times \text{Degree}
1 rad ≈ 57°16′; 1° ≈ 0.01746 rad1 रेडियन ≈ 57°16′; 1° ≈ 0.01746 रेडियन1 rad ≈ 57°16′; 1° ≈ 0.01746 rad
\text{Degree} = \frac{180}{\pi} \times \text{Radian}\text{अंश} = \frac{180}{\pi} \times \text{रेडियन}\text{Degree} = \frac{180}{\pi} \times \text{Radian}

Common values: 30° = \frac{\pi}{6}, 45° = \frac{\pi}{4}, 60° = \frac{\pi}{3}, 90° = \frac{\pi}{2}, 180° = π, 270° = \frac{3\pi}{2}, 360° = 2π. When no unit is written, the angle is in radians: θ = π/4 means π/4 radian.सामान्य मान: 30° = \frac{\pi}{6}, 45° = \frac{\pi}{4}, 60° = \frac{\pi}{3}, 90° = \frac{\pi}{2}, 180° = π, 270° = \frac{3\pi}{2}, 360° = 2π। इकाई न लिखी हो तो कोण रेडियन में समझें: θ = π/4 अर्थात् π/4 रेडियन।Common values: 30° = \frac{\pi}{6}, 45° = \frac{\pi}{4}, 60° = \frac{\pi}{3}, 90° = \frac{\pi}{2}, 180° = π, 270° = \frac{3\pi}{2}, 360° = 2π. Unit na likhi ho to angle radian mein samjhiye: θ = π/4 matlab π/4 radian.

Example 1Convert 40°20′ into radian measure.उदाहरण 140°20′ को रेडियन माप में बदलिए।Example 140°20′ ko radian measure mein badaliye.
Solution — step by step (2)हल — चरण दर चरण (2)Hal — step by step (2)
  1. 40^\circ 20' = 40\tfrac{1}{3}^\circ = \frac{121}{3}^\circ
  2. \frac{121}{3} \times \frac{\pi}{180} = \frac{121\pi}{540} \text{ radian}

3.3Trigonometric Functionsत्रिकोणमितीय फलनTrigonometric Functions

Take the unit circle (centre O, radius 1). Go from A(1, 0) along the circle through an arc of length x to the point P(a, b). We define \cos x = a and \sin x = b — this works for every real x, not only acute angles. Tap P in the unit circle picture.इकाई वृत्त लीजिए (केंद्र O, त्रिज्या 1)। A(1, 0) से वृत्त पर लंबाई x का चाप चलकर बिंदु P(a, b) पर पहुँचिए। परिभाषा: \cos x = a और \sin x = b — यह हर वास्तविक x के लिए है, केवल न्यून कोणों के लिए नहीं। इकाई वृत्त चित्र में P पर टैप कीजिए।Unit circle lijiye (centre O, radius 1). A(1, 0) se circle par lambai x ka arc chal kar point P(a, b) par pahunchiye. Definition: \cos x = a aur \sin x = b — ye har real x ke liye hai, sirf acute angles ke liye nahi. Unit circle picture mein P par tap kijiye.

Unit circle: cos x and sin x, and signs in the four quadrantsइकाई वृत्त: cos x और sin x, तथा चारों चतुर्थांशों में चिह्नUnit circle: cos x aur sin x, aur chaaron quadrants mein signsFig 3.6 · 9 tappable partsचित्र 3.6 · 9 टैप करने लायक हिस्सेFig 3.6 · 9 tappable partsOpen
\cos^2 x + \sin^2 x = 1
Pythagorean identitiesपाइथागोरस सर्वसमिकाएँPythagorean identities
1 + \tan^2 x = \sec^2 x
1 + \cot^2 x = \operatorname{cosec}^2 x
\tan x = \frac{\sin x}{\cos x}
Other four functions (where the denominator ≠ 0)अन्य चार फलन (जहाँ हर ≠ 0)Baaki chaar functions (jahan denominator ≠ 0)
\cot x = \frac{\cos x}{\sin x}
\sec x = \frac{1}{\cos x}
\operatorname{cosec} x = \frac{1}{\sin x}

Zeros: \sin x = 0 \iff x = n\pi; \cos x = 0 \iff x = (2n + 1)\frac{\pi}{2}, n ∈ Z. Periodicity: \sin(2n\pi + x) = \sin x, \cos(2n\pi + x) = \cos x. Negative angles: \sin(-x) = -\sin x, \cos(-x) = \cos x.शून्यक: \sin x = 0 \iff x = n\pi; \cos x = 0 \iff x = (2n + 1)\frac{\pi}{2}, n ∈ Z। आवर्तिता: \sin(2n\pi + x) = \sin x, \cos(2n\pi + x) = \cos x। ऋण कोण: \sin(-x) = -\sin x, \cos(-x) = \cos x।Zeros: \sin x = 0 \iff x = n\pi; \cos x = 0 \iff x = (2n + 1)\frac{\pi}{2}, n ∈ Z. Periodicity: \sin(2n\pi + x) = \sin x, \cos(2n\pi + x) = \cos x. Negative angles: \sin(-x) = -\sin x, \cos(-x) = \cos x.

Signs in quadrantsI: all positive; II: sin, cosec positive; III: tan, cot positive; IV: cos, sec positive ("All Silver Tea Cups"). Domains: sin, cos — R; tan, sec — \mathbf{R} - \{(2n+1)\frac{\pi}{2}\}; cot, cosec — \mathbf{R} - \{n\pi\}. Ranges: sin, cos — [−1, 1]; tan, cot — R; sec, cosec — \mathbf{R} - (-1, 1).चतुर्थांशों में चिह्नI: सभी धन; II: sin, cosec धन; III: tan, cot धन; IV: cos, sec धन ("All Silver Tea Cups")। प्रांत: sin, cos — R; tan, sec — \mathbf{R} - \{(2n+1)\frac{\pi}{2}\}; cot, cosec — \mathbf{R} - \{n\pi\}। परिसर: sin, cos — [−1, 1]; tan, cot — R; sec, cosec — \mathbf{R} - (-1, 1)।Quadrants mein signsI: sab positive; II: sin, cosec positive; III: tan, cot positive; IV: cos, sec positive ("All Silver Tea Cups"). Domains: sin, cos — R; tan, sec — \mathbf{R} - \{(2n+1)\frac{\pi}{2}\}; cot, cosec — \mathbf{R} - \{n\pi\}. Ranges: sin, cos — [−1, 1]; tan, cot — R; sec, cosec — \mathbf{R} - (-1, 1).

The graphs of sin and cos are waves repeating every 2π — tap them in the sin–cos graph; tan and cot repeat every π.sin और cos के आलेख तरंगें हैं जो हर 2π पर दोहराती हैं — sin–cos आलेख में टैप कीजिए; tan और cot हर π पर दोहराते हैं।sin aur cos ke graphs waves hain jo har 2π par repeat hoti hain — sin–cos graph mein tap kijiye; tan aur cot har π par repeat hote hain.

Graphs of y = sin x and y = cos xy = sin x और y = cos x के आलेखy = sin x aur y = cos x ke graphsSection 3.3 · 3 tappable partsअनुभाग 3.3 · 3 टैप करने लायक हिस्सेSection 3.3 · 3 tappable partsOpen
Example 7If \cos x = -\frac{3}{5} and x lies in the third quadrant, find the other five trigonometric functions.उदाहरण 7यदि \cos x = -\frac{3}{5} और x तीसरे चतुर्थांश में है, तो अन्य पाँच त्रिकोणमितीय फलन ज्ञात कीजिए।Example 7Agar \cos x = -\frac{3}{5} aur x teesre quadrant mein hai, to baaki paanch trigonometric functions nikaliye.
Solution — step by step (2)हल — चरण दर चरण (2)Hal — step by step (2)
  1. \sin^2 x = 1 - \frac{9}{25} = \frac{16}{25} \Rightarrow \sin x = -\frac{4}{5} \ (\text{III: } \sin < 0)
  2. \sec x = -\frac{5}{3},\ \operatorname{cosec} x = -\frac{5}{4},\ \tan x = \frac{4}{3},\ \cot x = \frac{3}{4}
Example 10Find the value of \sin\frac{31\pi}{3}.उदाहरण 10\sin\frac{31\pi}{3} का मान ज्ञात कीजिए।Example 10\sin\frac{31\pi}{3} ki value nikaliye.
Solution — step by step (1)हल — चरण दर चरण (1)Hal — step by step (1)
  1. \sin\frac{31\pi}{3} = \sin\left(10\pi + \frac{\pi}{3}\right) = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}

3.4Trigonometric Functions of Sum and Difference of Two Anglesदो कोणों के योग और अंतर के त्रिकोणमितीय फलनDo angles ke sum aur difference ke trigonometric functions

\cos(x \pm y) = \cos x \cos y \mp \sin x \sin y
Note the sign flips for cos.ध्यान दें: cos में चिह्न उलट जाता है।Dhyaan dijiye: cos mein sign ulat jaata hai.
\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y
Sine keeps the same sign.sin में वही चिह्न रहता है।sin mein wahi sign rehta hai.
\tan(x \pm y) = \frac{\tan x \pm \tan y}{1 \mp \tan x \tan y}
(none of x, y, x ± y an odd multiple of π/2)(x, y, x ± y में से कोई π/2 का विषम गुणज न हो)(x, y, x ± y mein se koi π/2 ka odd multiple na ho)
\cos\left(\tfrac{\pi}{2} - x\right) = \sin x
Allied anglesसंबद्ध कोणAllied angles
\sin\left(\tfrac{\pi}{2} - x\right) = \cos x
\cos(\pi \pm x) = -\cos x
\sin(\pi - x) = \sin x
\sin(\pi + x) = -\sin x
\sin 2x = 2\sin x \cos x = \frac{2\tan x}{1 + \tan^2 x}
Double angle (sine)द्विगुण कोण (sine)Double angle (sine)
\cos 2x = \cos^2 x - \sin^2 x = 2\cos^2 x - 1
Double angle (cosine) — four formsद्विगुण कोण (cosine) — चार रूपDouble angle (cosine) — chaar roop
\cos 2x = 1 - 2\sin^2 x = \frac{1 - \tan^2 x}{1 + \tan^2 x}
\tan 2x = \frac{2\tan x}{1 - \tan^2 x}
Double and triple anglesद्विगुण और त्रिगुण कोणDouble aur triple angles
\sin 3x = 3\sin x - 4\sin^3 x
\cos 3x = 4\cos^3 x - 3\cos x
\tan 3x = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x}
\cos x + \cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2}
Sum-to-product (cos)योग से गुणनफल (cos)Sum-to-product (cos)
\cos x - \cos y = -2\sin\frac{x+y}{2}\sin\frac{x-y}{2}
\sin x + \sin y = 2\sin\frac{x+y}{2}\cos\frac{x-y}{2}
Sum-to-product (sin)योग से गुणनफल (sin)Sum-to-product (sin)
\sin x - \sin y = 2\cos\frac{x+y}{2}\sin\frac{x-y}{2}
2\cos x \cos y = \cos(x+y) + \cos(x-y)
Product-to-sumगुणनफल से योगProduct-to-sum
-2\sin x \sin y = \cos(x+y) - \cos(x-y)
2\sin x \cos y = \sin(x+y) + \sin(x-y)
Example 12Find the value of \sin 15^\circ.उदाहरण 12\sin 15^\circ का मान ज्ञात कीजिए।Example 12\sin 15^\circ ki value nikaliye.
Solution — step by step (2)हल — चरण दर चरण (2)Hal — step by step (2)
  1. \sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ
  2. = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} - 1}{2\sqrt{2}}
Example 17Prove that \frac{\sin 5x - 2\sin 3x + \sin x}{\cos 5x - \cos x} = \tan x.उदाहरण 17सिद्ध कीजिए कि \frac{\sin 5x - 2\sin 3x + \sin x}{\cos 5x - \cos x} = \tan x।Example 17Prove kijiye ki \frac{\sin 5x - 2\sin 3x + \sin x}{\cos 5x - \cos x} = \tan x.
Solution — step by step (3)हल — चरण दर चरण (3)Hal — step by step (3)
  1. \sin 5x + \sin x = 2\sin 3x \cos 2x \Rightarrow \text{Num} = 2\sin 3x(\cos 2x - 1) = -4\sin 3x \sin^2 x
  2. \cos 5x - \cos x = -2\sin 3x \sin 2x = -4\sin 3x \sin x \cos x
  3. \frac{-4\sin 3x \sin^2 x}{-4\sin 3x \sin x \cos x} = \frac{\sin x}{\cos x} = \tan x

★ Yaad rakho — key formulas★ याद रखो — मुख्य सूत्र★ Yaad rakho — key formulas

  1. \pi rad = 180°; 1 rad ≈ 57°16′; l = r\theta (θ in radian).\pi रेडियन = 180°; 1 रेडियन ≈ 57°16′; l = r\theta (θ रेडियन में)।\pi radian = 180°; 1 radian ≈ 57°16′; l = r\theta (θ radian mein).
  2. ASTC: All positive in I, Sin (cosec) in II, Tan (cot) in III, Cos (sec) in IV.ASTC: I में सभी धनात्मक, II में sin (cosec), III में tan (cot), IV में cos (sec)।ASTC: I mein sab positive, II mein sin (cosec), III mein tan (cot), IV mein cos (sec).
  3. sin, cos have period 2\pi; tan, cot have period \pi.sin, cos का आवर्तकाल 2\pi; tan, cot का \pi।sin, cos ka period 2\pi; tan, cot ka \pi.
  4. \sin(-x) = -\sin x, \cos(-x) = \cos x, \tan(-x) = -\tan x.\sin(-x) = -\sin x, \cos(-x) = \cos x, \tan(-x) = -\tan x।\sin(-x) = -\sin x, \cos(-x) = \cos x, \tan(-x) = -\tan x.
  5. \sin 2x = 2\sin x\cos x; \cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1; \tan 2x = \frac{2\tan x}{1 - \tan^2 x}.\sin 2x = 2\sin x\cos x; \cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1; \tan 2x = \frac{2\tan x}{1 - \tan^2 x}।\sin 2x = 2\sin x\cos x; \cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1; \tan 2x = \frac{2\tan x}{1 - \tan^2 x}.
  6. Half angle: \sin\frac{x}{2} = \pm\sqrt{\frac{1 - \cos x}{2}}, \cos\frac{x}{2} = \pm\sqrt{\frac{1 + \cos x}{2}} — sign from the quadrant of \frac{x}{2}.अर्ध कोण: \sin\frac{x}{2} = \pm\sqrt{\frac{1 - \cos x}{2}}, \cos\frac{x}{2} = \pm\sqrt{\frac{1 + \cos x}{2}} — चिह्न \frac{x}{2} के चतुर्थांश से।Half angle: \sin\frac{x}{2} = \pm\sqrt{\frac{1 - \cos x}{2}}, \cos\frac{x}{2} = \pm\sqrt{\frac{1 + \cos x}{2}} — sign \frac{x}{2} ke quadrant se.

⚠ Common mistakes⚠ आम गलतियाँ⚠ Common mistakes

  • Using l = r\theta with θ in degrees. θ must be in radian.l = r\theta में θ डिग्री में रखना। θ रेडियन में होना चाहिए।l = r\theta mein θ degree mein rakhna. θ radian mein hona chahiye.
  • Taking the sign from the quadrant of x when finding \sin\frac{x}{2}. Use the quadrant of \frac{x}{2}.\sin\frac{x}{2} निकालते समय x के चतुर्थांश से चिह्न लेना। \frac{x}{2} का चतुर्थांश देखिए।\sin\frac{x}{2} nikalte waqt x ke quadrant se sign lena. \frac{x}{2} ka quadrant dekhiye.
  • \cos(x + y) = \cos x + \cos y — wrong; trig functions are not linear.\cos(x + y) = \cos x + \cos y — गलत; त्रिकोणमितीय फलन रैखिक नहीं हैं।\cos(x + y) = \cos x + \cos y — galat; trig functions linear nahi hain.
  • Writing \cos(x - y) = \cos x\cos y - \sin x\sin y. For cos the sign flips: it is +.\cos(x - y) = \cos x\cos y - \sin x\sin y लिखना। cos में चिह्न उलटता है: यह + है।\cos(x - y) = \cos x\cos y - \sin x\sin y likhna. cos mein sign ulat-ta hai: ye + hai.
  • Reducing big angles wrongly: subtract multiples of 360° (2π) for sin/cos, but 180° (π) is enough for tan/cot.बड़े कोण गलत घटाना: sin/cos के लिए 360° (2π) के गुणज घटाइए, tan/cot के लिए 180° (π) काफी है।Bade angles galat ghatana: sin/cos ke liye 360° (2π) ke multiples ghataiye, tan/cot ke liye 180° (π) kaafi hai.

✎ NCERT exercises — solutions✎ NCERT प्रश्नावली — हल✎ NCERT exercises — solutions

All 52 questions of the NCERT exercises. Tap a question to see the solution — try it yourself first!NCERT प्रश्नावली के सभी 52 प्रश्न। हल देखने के लिए प्रश्न पर टैप करें — पहले स्वयं हल करके देखें!NCERT exercises ke saare 52 questions. Hal dekhne ke liye question par tap kijiye — pehle khud try kijiye!

Exercise 3.1प्रश्नावली 3.1Exercise 3.1 · 7 Q

Q1. Find the radian measures of (i) 25° (ii) −47°30′ (iii) 240° (iv) 520°.रेडियन माप ज्ञात कीजिए: (i) 25° (ii) −47°30′ (iii) 240° (iv) 520°।Find the radian measures of (i) 25° (ii) −47°30′ (iii) 240° (iv) 520°.
  1. (i)\ 25 \times \frac{\pi}{180} = \frac{5\pi}{36} \qquad (ii)\ -47\tfrac{1}{2} \times \frac{\pi}{180} = -\frac{95}{2} \cdot \frac{\pi}{180} = -\frac{19\pi}{72}
  2. (iii)\ 240 \times \frac{\pi}{180} = \frac{4\pi}{3} \qquad (iv)\ 520 \times \frac{\pi}{180} = \frac{26\pi}{9}
Q2. Find the degree measures (use \pi = \frac{22}{7}): (i) \frac{11}{16} (ii) −4 (iii) \frac{5\pi}{3} (iv) \frac{7\pi}{6}.डिग्री माप ज्ञात कीजिए (\pi = \frac{22}{7} लीजिए): (i) \frac{11}{16} (ii) −4 (iii) \frac{5\pi}{3} (iv) \frac{7\pi}{6}।Find the degree measures (use \pi = \frac{22}{7}): (i) \frac{11}{16} (ii) −4 (iii) \frac{5\pi}{3} (iv) \frac{7\pi}{6}.
  1. (i)\ \frac{11}{16} \times \frac{180 \times 7}{22} = \frac{315}{8}^\circ = 39^\circ 22' 30''
  2. (ii)\ -4 \times \frac{180 \times 7}{22} = -\frac{2520}{11}^\circ = -229^\circ 5' 27'' \text{ (approx.)}
  3. (iii)\ \frac{5\pi}{3} = \frac{5 \times 180^\circ}{3} = 300^\circ \qquad (iv)\ \frac{7\pi}{6} = 210^\circ
Q3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?एक पहिया एक मिनट में 360 परिक्रमण करता है। एक सेकंड में यह कितने रेडियन घूमता है?A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
  1. 360 \text{ rev/min} = 6 \text{ rev/s} = 6 \times 2\pi = 12\pi \text{ radian}360 \text{ चक्कर/मिनट} = 6 \text{ चक्कर/सेकंड} = 6 \times 2\pi = 12\pi \text{ रेडियन}360 \text{ chakkar/min} = 6 \text{ chakkar/s} = 6 \times 2\pi = 12\pi \text{ radian}
Q4. Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of 22 cm (\pi = \frac{22}{7}).100 cm त्रिज्या वाले वृत्त में 22 cm लंबाई का चाप केंद्र पर कितने डिग्री का कोण बनाता है? (\pi = \frac{22}{7})Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of 22 cm (\pi = \frac{22}{7}).
  1. \theta = \frac{l}{r} = \frac{22}{100} \text{ rad} = \frac{22}{100} \times \frac{180 \times 7}{22} = 12.6^\circ = 12^\circ 36'
Q5. In a circle of diameter 40 cm a chord is 20 cm long. Find the length of the minor arc of the chord.40 cm व्यास के वृत्त में एक जीवा 20 cm लंबी है। जीवा के लघु चाप की लंबाई ज्ञात कीजिए।In a circle of diameter 40 cm a chord is 20 cm long. Find the length of the minor arc of the chord.
  1. r = 20 cm and chord = 20 cm, so the triangle formed with the centre is equilateral: angle = 60° = π/3.r = 20 cm और जीवा = 20 cm, अतः केंद्र के साथ बना त्रिभुज समबाहु है: कोण = 60° = π/3।r = 20 cm aur chord = 20 cm, isliye centre ke saath bana triangle equilateral hai: angle = 60° = π/3.
    l = r\theta = 20 \times \frac{\pi}{3} = \frac{20\pi}{3} \text{ cm}
Q6. Arcs of the same length subtend 60° and 75° at the centres of two circles. Find the ratio of their radii.दो वृत्तों में समान लंबाई के चाप केंद्र पर 60° और 75° बनाते हैं। उनकी त्रिज्याओं का अनुपात ज्ञात कीजिए।Arcs of the same length subtend 60° and 75° at the centres of two circles. Find the ratio of their radii.
  1. r_1 \theta_1 = r_2 \theta_2 \Rightarrow r_1 \cdot \frac{\pi}{3} = r_2 \cdot \frac{5\pi}{12} \Rightarrow r_1 : r_2 = 5 : 4
Q7. A pendulum 75 cm long; find the angle (radian) it swings if the tip describes an arc of (i) 10 cm (ii) 15 cm (iii) 21 cm.75 cm लंबा लोलक; यदि सिरा (i) 10 cm (ii) 15 cm (iii) 21 cm का चाप बनाए तो कोण (रेडियन) ज्ञात कीजिए।A pendulum 75 cm long; find the angle (radian) it swings if the tip describes an arc of (i) 10 cm (ii) 15 cm (iii) 21 cm.
  1. \theta = \frac{l}{75}:\quad (i)\ \frac{10}{75} = \frac{2}{15} \quad (ii)\ \frac{15}{75} = \frac{1}{5} \quad (iii)\ \frac{21}{75} = \frac{7}{25}

Exercise 3.2प्रश्नावली 3.2Exercise 3.2 · 10 Q

Q1. \cos x = -\frac{1}{2}, x in quadrant III. Find the other five.\cos x = -\frac{1}{2}, x तीसरे चतुर्थांश में। शेष पाँच फलन ज्ञात कीजिए।\cos x = -\frac{1}{2}, x in quadrant III. Find the other five.
  1. \sin x = -\sqrt{1 - \tfrac{1}{4}} = -\frac{\sqrt{3}}{2} \ (\text{III: sin} < 0)
  2. \operatorname{cosec} x = -\frac{2}{\sqrt{3}},\ \sec x = -2,\ \tan x = \frac{\sin x}{\cos x} = \sqrt{3},\ \cot x = \frac{1}{\sqrt{3}}
Q2. \sin x = \frac{3}{5}, x in quadrant II.\sin x = \frac{3}{5}, x दूसरे चतुर्थांश में।\sin x = \frac{3}{5}, x in quadrant II.
  1. \cos x = -\frac{4}{5} \ (\text{II: cos} < 0),\ \operatorname{cosec} x = \frac{5}{3},\ \sec x = -\frac{5}{4},\ \tan x = -\frac{3}{4},\ \cot x = -\frac{4}{3}
Q3. \cot x = \frac{3}{4}, x in quadrant III.\cot x = \frac{3}{4}, x तीसरे चतुर्थांश में।\cot x = \frac{3}{4}, x in quadrant III.
  1. \tan x = \frac{4}{3},\ \sec^2 x = 1 + \frac{16}{9} = \frac{25}{9} \Rightarrow \sec x = -\frac{5}{3} \ (\text{III})
  2. \cos x = -\frac{3}{5},\ \sin x = \tan x \cos x = -\frac{4}{5},\ \operatorname{cosec} x = -\frac{5}{4}
Q4. \sec x = \frac{13}{5}, x in quadrant IV.\sec x = \frac{13}{5}, x चौथे चतुर्थांश में।\sec x = \frac{13}{5}, x in quadrant IV.
  1. \cos x = \frac{5}{13},\ \sin x = -\frac{12}{13} \ (\text{IV}),\ \operatorname{cosec} x = -\frac{13}{12},\ \tan x = -\frac{12}{5},\ \cot x = -\frac{5}{12}
Q5. \tan x = -\frac{5}{12}, x in quadrant II.\tan x = -\frac{5}{12}, x दूसरे चतुर्थांश में।\tan x = -\frac{5}{12}, x in quadrant II.
  1. \sec^2 x = 1 + \frac{25}{144} = \frac{169}{144} \Rightarrow \sec x = -\frac{13}{12} \ (\text{II}),\ \cos x = -\frac{12}{13}
  2. \sin x = \tan x \cos x = \frac{5}{13},\ \operatorname{cosec} x = \frac{13}{5},\ \cot x = -\frac{12}{5}
Q6. Find sin 765°.sin 765° ज्ञात कीजिए।Find sin 765°.
  1. \sin 765^\circ = \sin(2 \times 360^\circ + 45^\circ) = \sin 45^\circ = \frac{1}{\sqrt{2}}
Q7. Find cosec (−1410°).cosec (−1410°) ज्ञात कीजिए।Find cosec (−1410°).
  1. \operatorname{cosec}(-1410^\circ) = \operatorname{cosec}(-1440^\circ + 30^\circ) = \operatorname{cosec}(-4 \times 360^\circ + 30^\circ) = \operatorname{cosec} 30^\circ = 2
Q8. Find \tan\frac{19\pi}{3}.\tan\frac{19\pi}{3} ज्ञात कीजिए।Find \tan\frac{19\pi}{3}.
  1. \tan\frac{19\pi}{3} = \tan\left(6\pi + \frac{\pi}{3}\right) = \tan\frac{\pi}{3} = \sqrt{3}
Q9. Find \sin\left(-\frac{11\pi}{3}\right).\sin\left(-\frac{11\pi}{3}\right) ज्ञात कीजिए।Find \sin\left(-\frac{11\pi}{3}\right).
  1. \sin\left(-\frac{11\pi}{3}\right) = \sin\left(-4\pi + \frac{\pi}{3}\right) = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}
Q10. Find \cot\left(-\frac{15\pi}{4}\right).\cot\left(-\frac{15\pi}{4}\right) ज्ञात कीजिए।Find \cot\left(-\frac{15\pi}{4}\right).
  1. \cot\left(-\frac{15\pi}{4}\right) = \cot\left(-4\pi + \frac{\pi}{4}\right) = \cot\frac{\pi}{4} = 1

Exercise 3.3प्रश्नावली 3.3Exercise 3.3 · 25 Q

Q1. Prove \sin^2\frac{\pi}{6} + \cos^2\frac{\pi}{3} - \tan^2\frac{\pi}{4} = -\frac{1}{2}.सिद्ध कीजिए \sin^2\frac{\pi}{6} + \cos^2\frac{\pi}{3} - \tan^2\frac{\pi}{4} = -\frac{1}{2}।Prove \sin^2\frac{\pi}{6} + \cos^2\frac{\pi}{3} - \tan^2\frac{\pi}{4} = -\frac{1}{2}.
  1. \frac{1}{4} + \frac{1}{4} - 1 = -\frac{1}{2}
Q2. Prove 2\sin^2\frac{\pi}{6} + \operatorname{cosec}^2\frac{7\pi}{6}\cos^2\frac{\pi}{3} = \frac{3}{2}.सिद्ध कीजिए 2\sin^2\frac{\pi}{6} + \operatorname{cosec}^2\frac{7\pi}{6}\cos^2\frac{\pi}{3} = \frac{3}{2}।Prove 2\sin^2\frac{\pi}{6} + \operatorname{cosec}^2\frac{7\pi}{6}\cos^2\frac{\pi}{3} = \frac{3}{2}.
  1. \operatorname{cosec}\frac{7\pi}{6} = -\operatorname{cosec}\frac{\pi}{6} = -2 \Rightarrow 2 \cdot \frac{1}{4} + 4 \cdot \frac{1}{4} = \frac{1}{2} + 1 = \frac{3}{2}
Q3. Prove \cot^2\frac{\pi}{6} + \operatorname{cosec}\frac{5\pi}{6} + 3\tan^2\frac{\pi}{6} = 6.सिद्ध कीजिए \cot^2\frac{\pi}{6} + \operatorname{cosec}\frac{5\pi}{6} + 3\tan^2\frac{\pi}{6} = 6।Prove \cot^2\frac{\pi}{6} + \operatorname{cosec}\frac{5\pi}{6} + 3\tan^2\frac{\pi}{6} = 6.
  1. 3 + \operatorname{cosec}\left(\pi - \frac{\pi}{6}\right) + 3 \cdot \frac{1}{3} = 3 + 2 + 1 = 6
Q4. Prove 2\sin^2\frac{3\pi}{4} + 2\cos^2\frac{\pi}{4} + 2\sec^2\frac{\pi}{3} = 10.सिद्ध कीजिए 2\sin^2\frac{3\pi}{4} + 2\cos^2\frac{\pi}{4} + 2\sec^2\frac{\pi}{3} = 10।Prove 2\sin^2\frac{3\pi}{4} + 2\cos^2\frac{\pi}{4} + 2\sec^2\frac{\pi}{3} = 10.
  1. 2 \cdot \frac{1}{2} + 2 \cdot \frac{1}{2} + 2 \cdot 4 = 1 + 1 + 8 = 10
Q5. Find (i) sin 75° (ii) tan 15°.ज्ञात कीजिए (i) sin 75° (ii) tan 15°।Find (i) sin 75° (ii) tan 15°.
  1. (i)\ \sin(45^\circ + 30^\circ) = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}}
  2. (ii)\ \tan(45^\circ - 30^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} - 1)^2}{2} = 2 - \sqrt{3}
Q6. Prove \cos\left(\frac{\pi}{4} - x\right)\cos\left(\frac{\pi}{4} - y\right) - \sin\left(\frac{\pi}{4} - x\right)\sin\left(\frac{\pi}{4} - y\right) = \sin(x + y).सिद्ध कीजिए \cos\left(\frac{\pi}{4} - x\right)\cos\left(\frac{\pi}{4} - y\right) - \sin\left(\frac{\pi}{4} - x\right)\sin\left(\frac{\pi}{4} - y\right) = \sin(x + y)।Prove \cos\left(\frac{\pi}{4} - x\right)\cos\left(\frac{\pi}{4} - y\right) - \sin\left(\frac{\pi}{4} - x\right)\sin\left(\frac{\pi}{4} - y\right) = \sin(x + y).
  1. \text{LHS} = \cos\left(\frac{\pi}{4} - x + \frac{\pi}{4} - y\right) = \cos\left(\frac{\pi}{2} - (x + y)\right) = \sin(x + y)
Q7. Prove \frac{\tan(\frac{\pi}{4} + x)}{\tan(\frac{\pi}{4} - x)} = \left(\frac{1 + \tan x}{1 - \tan x}\right)^2.सिद्ध कीजिए \frac{\tan(\frac{\pi}{4} + x)}{\tan(\frac{\pi}{4} - x)} = \left(\frac{1 + \tan x}{1 - \tan x}\right)^2।Prove \frac{\tan(\frac{\pi}{4} + x)}{\tan(\frac{\pi}{4} - x)} = \left(\frac{1 + \tan x}{1 - \tan x}\right)^2.
  1. \tan\left(\frac{\pi}{4} + x\right) = \frac{1 + \tan x}{1 - \tan x},\quad \tan\left(\frac{\pi}{4} - x\right) = \frac{1 - \tan x}{1 + \tan x}
  2. \text{Ratio} = \frac{1 + \tan x}{1 - \tan x} \cdot \frac{1 + \tan x}{1 - \tan x} = \left(\frac{1 + \tan x}{1 - \tan x}\right)^2
Q8. Prove \frac{\cos(\pi + x)\cos(-x)}{\sin(\pi - x)\cos(\frac{\pi}{2} + x)} = \cot^2 x.सिद्ध कीजिए \frac{\cos(\pi + x)\cos(-x)}{\sin(\pi - x)\cos(\frac{\pi}{2} + x)} = \cot^2 x।Prove \frac{\cos(\pi + x)\cos(-x)}{\sin(\pi - x)\cos(\frac{\pi}{2} + x)} = \cot^2 x.
  1. \frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)} = \frac{\cos^2 x}{\sin^2 x} = \cot^2 x
Q9. Prove \cos\left(\frac{3\pi}{2} + x\right)\cos(2\pi + x)\left[\cot\left(\frac{3\pi}{2} - x\right) + \cot(2\pi + x)\right] = 1.सिद्ध कीजिए \cos\left(\frac{3\pi}{2} + x\right)\cos(2\pi + x)\left[\cot\left(\frac{3\pi}{2} - x\right) + \cot(2\pi + x)\right] = 1।Prove \cos\left(\frac{3\pi}{2} + x\right)\cos(2\pi + x)\left[\cot\left(\frac{3\pi}{2} - x\right) + \cot(2\pi + x)\right] = 1.
  1. \cos\left(\tfrac{3\pi}{2} + x\right) = \sin x,\ \cos(2\pi + x) = \cos x,\ \cot\left(\tfrac{3\pi}{2} - x\right) = \tan x,\ \cot(2\pi + x) = \cot x
  2. \sin x \cos x(\tan x + \cot x) = \sin x\cos x \cdot \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = 1
Q10. Prove \sin(n+1)x\,\sin(n+2)x + \cos(n+1)x\,\cos(n+2)x = \cos x.सिद्ध कीजिए \sin(n+1)x\,\sin(n+2)x + \cos(n+1)x\,\cos(n+2)x = \cos x।Prove \sin(n+1)x\,\sin(n+2)x + \cos(n+1)x\,\cos(n+2)x = \cos x.
  1. \text{LHS} = \cos[(n+2)x - (n+1)x] = \cos x
Q11. Prove \cos\left(\frac{3\pi}{4} + x\right) - \cos\left(\frac{3\pi}{4} - x\right) = -\sqrt{2}\sin x.सिद्ध कीजिए \cos\left(\frac{3\pi}{4} + x\right) - \cos\left(\frac{3\pi}{4} - x\right) = -\sqrt{2}\sin x।Prove \cos\left(\frac{3\pi}{4} + x\right) - \cos\left(\frac{3\pi}{4} - x\right) = -\sqrt{2}\sin x.
  1. = -2\sin\frac{3\pi}{4}\sin x = -2 \cdot \frac{1}{\sqrt{2}}\sin x = -\sqrt{2}\sin x
Q12. Prove \sin^2 6x - \sin^2 4x = \sin 2x \sin 10x.सिद्ध कीजिए \sin^2 6x - \sin^2 4x = \sin 2x \sin 10x।Prove \sin^2 6x - \sin^2 4x = \sin 2x \sin 10x.
  1. (\sin 6x + \sin 4x)(\sin 6x - \sin 4x) = (2\sin 5x\cos x)(2\cos 5x \sin x)
  2. = (2\sin 5x\cos 5x)(2\sin x\cos x) = \sin 10x \sin 2x
Q13. Prove \cos^2 2x - \cos^2 6x = \sin 4x \sin 8x.सिद्ध कीजिए \cos^2 2x - \cos^2 6x = \sin 4x \sin 8x।Prove \cos^2 2x - \cos^2 6x = \sin 4x \sin 8x.
  1. (\cos 2x + \cos 6x)(\cos 2x - \cos 6x) = (2\cos 4x\cos 2x)(2\sin 4x \sin 2x)
  2. = (2\sin 4x\cos 4x)(2\sin 2x\cos 2x) = \sin 8x \sin 4x
Q14. Prove \sin 2x + 2\sin 4x + \sin 6x = 4\cos^2 x \sin 4x.सिद्ध कीजिए \sin 2x + 2\sin 4x + \sin 6x = 4\cos^2 x \sin 4x।Prove \sin 2x + 2\sin 4x + \sin 6x = 4\cos^2 x \sin 4x.
  1. \sin 6x + \sin 2x = 2\sin 4x\cos 2x \Rightarrow \text{LHS} = 2\sin 4x(\cos 2x + 1) = 2\sin 4x \cdot 2\cos^2 x
Q15. Prove \cot 4x(\sin 5x + \sin 3x) = \cot x(\sin 5x - \sin 3x).सिद्ध कीजिए \cot 4x(\sin 5x + \sin 3x) = \cot x(\sin 5x - \sin 3x)।Prove \cot 4x(\sin 5x + \sin 3x) = \cot x(\sin 5x - \sin 3x).
  1. \text{LHS} = \frac{\cos 4x}{\sin 4x} \cdot 2\sin 4x\cos x = 2\cos 4x\cos x
  2. \text{RHS} = \frac{\cos x}{\sin x} \cdot 2\cos 4x\sin x = 2\cos 4x\cos x
Q16. Prove \frac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x} = -\frac{\sin 2x}{\cos 10x}.सिद्ध कीजिए \frac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x} = -\frac{\sin 2x}{\cos 10x}।Prove \frac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x} = -\frac{\sin 2x}{\cos 10x}.
  1. \frac{-2\sin 7x\sin 2x}{2\cos 10x\sin 7x} = -\frac{\sin 2x}{\cos 10x}
Q17. Prove \frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x.सिद्ध कीजिए \frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x।Prove \frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x.
  1. \frac{2\sin 4x\cos x}{2\cos 4x\cos x} = \tan 4x
Q18. Prove \frac{\sin x - \sin y}{\cos x + \cos y} = \tan\frac{x - y}{2}.सिद्ध कीजिए \frac{\sin x - \sin y}{\cos x + \cos y} = \tan\frac{x - y}{2}।Prove \frac{\sin x - \sin y}{\cos x + \cos y} = \tan\frac{x - y}{2}.
  1. \frac{2\cos\frac{x+y}{2}\sin\frac{x-y}{2}}{2\cos\frac{x+y}{2}\cos\frac{x-y}{2}} = \tan\frac{x - y}{2}
Q19. Prove \frac{\sin x + \sin 3x}{\cos x + \cos 3x} = \tan 2x.सिद्ध कीजिए \frac{\sin x + \sin 3x}{\cos x + \cos 3x} = \tan 2x।Prove \frac{\sin x + \sin 3x}{\cos x + \cos 3x} = \tan 2x.
  1. \frac{2\sin 2x\cos x}{2\cos 2x\cos x} = \tan 2x
Q20. Prove \frac{\sin x - \sin 3x}{\sin^2 x - \cos^2 x} = 2\sin x.सिद्ध कीजिए \frac{\sin x - \sin 3x}{\sin^2 x - \cos^2 x} = 2\sin x।Prove \frac{\sin x - \sin 3x}{\sin^2 x - \cos^2 x} = 2\sin x.
  1. \frac{-2\cos 2x \sin x}{-\cos 2x} = 2\sin x
Q21. Prove \frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x.सिद्ध कीजिए \frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x।Prove \frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x.
  1. \frac{2\cos 3x\cos x + \cos 3x}{2\sin 3x\cos x + \sin 3x} = \frac{\cos 3x(2\cos x + 1)}{\sin 3x(2\cos x + 1)} = \cot 3x
Q22. Prove \cot x\cot 2x - \cot 2x\cot 3x - \cot 3x\cot x = 1.सिद्ध कीजिए \cot x\cot 2x - \cot 2x\cot 3x - \cot 3x\cot x = 1।Prove \cot x\cot 2x - \cot 2x\cot 3x - \cot 3x\cot x = 1.
  1. \cot 3x = \cot(x + 2x) = \frac{\cot x\cot 2x - 1}{\cot x + \cot 2x}
  2. \Rightarrow \cot 3x(\cot x + \cot 2x) = \cot x\cot 2x - 1 \Rightarrow \cot x\cot 2x - \cot 2x\cot 3x - \cot 3x\cot x = 1
Q23. Prove \tan 4x = \frac{4\tan x(1 - \tan^2 x)}{1 - 6\tan^2 x + \tan^4 x}.सिद्ध कीजिए \tan 4x = \frac{4\tan x(1 - \tan^2 x)}{1 - 6\tan^2 x + \tan^4 x}।Prove \tan 4x = \frac{4\tan x(1 - \tan^2 x)}{1 - 6\tan^2 x + \tan^4 x}.
  1. \text{Let } t = \tan x,\ \tan 2x = \frac{2t}{1 - t^2}
  2. \tan 4x = \frac{2\tan 2x}{1 - \tan^2 2x} = \frac{\frac{4t}{1 - t^2}}{1 - \frac{4t^2}{(1 - t^2)^2}} = \frac{4t(1 - t^2)}{(1 - t^2)^2 - 4t^2} = \frac{4t(1 - t^2)}{1 - 6t^2 + t^4}
Q24. Prove \cos 4x = 1 - 8\sin^2 x\cos^2 x.सिद्ध कीजिए \cos 4x = 1 - 8\sin^2 x\cos^2 x।Prove \cos 4x = 1 - 8\sin^2 x\cos^2 x.
  1. \cos 4x = 1 - 2\sin^2 2x = 1 - 2(2\sin x\cos x)^2 = 1 - 8\sin^2 x\cos^2 x
Q25. Prove \cos 6x = 32\cos^6 x - 48\cos^4 x + 18\cos^2 x - 1.सिद्ध कीजिए \cos 6x = 32\cos^6 x - 48\cos^4 x + 18\cos^2 x - 1।Prove \cos 6x = 32\cos^6 x - 48\cos^4 x + 18\cos^2 x - 1.
  1. \cos 6x = \cos 3(2x) = 4\cos^3 2x - 3\cos 2x,\quad \cos 2x = 2c^2 - 1 \ (c = \cos x)
  2. = 4(2c^2 - 1)^3 - 3(2c^2 - 1) = 4(8c^6 - 12c^4 + 6c^2 - 1) - 6c^2 + 3
  3. = 32c^6 - 48c^4 + 18c^2 - 1

Miscellaneous Exercise on Chapter 3अध्याय 3 पर विविध प्रश्नावलीMiscellaneous Exercise on Chapter 3 · 10 Q

Q1. Prove 2\cos\frac{\pi}{13}\cos\frac{9\pi}{13} + \cos\frac{3\pi}{13} + \cos\frac{5\pi}{13} = 0.सिद्ध कीजिए 2\cos\frac{\pi}{13}\cos\frac{9\pi}{13} + \cos\frac{3\pi}{13} + \cos\frac{5\pi}{13} = 0।Prove 2\cos\frac{\pi}{13}\cos\frac{9\pi}{13} + \cos\frac{3\pi}{13} + \cos\frac{5\pi}{13} = 0.
  1. 2\cos\frac{\pi}{13}\cos\frac{9\pi}{13} = \cos\frac{10\pi}{13} + \cos\frac{8\pi}{13}
  2. \cos\frac{10\pi}{13} = -\cos\frac{3\pi}{13},\ \cos\frac{8\pi}{13} = -\cos\frac{5\pi}{13} \Rightarrow \text{sum} = 0
Q2. Prove (\sin 3x + \sin x)\sin x + (\cos 3x - \cos x)\cos x = 0.सिद्ध कीजिए (\sin 3x + \sin x)\sin x + (\cos 3x - \cos x)\cos x = 0।Prove (\sin 3x + \sin x)\sin x + (\cos 3x - \cos x)\cos x = 0.
  1. 2\sin 2x\cos x \sin x - 2\sin 2x\sin x\cos x = 0
Q3. Prove (\cos x + \cos y)^2 + (\sin x - \sin y)^2 = 4\cos^2\frac{x + y}{2}.सिद्ध कीजिए (\cos x + \cos y)^2 + (\sin x - \sin y)^2 = 4\cos^2\frac{x + y}{2}।Prove (\cos x + \cos y)^2 + (\sin x - \sin y)^2 = 4\cos^2\frac{x + y}{2}.
  1. = 2 + 2(\cos x\cos y - \sin x\sin y) = 2 + 2\cos(x + y) = 2 \cdot 2\cos^2\frac{x + y}{2}
Q4. Prove (\cos x - \cos y)^2 + (\sin x - \sin y)^2 = 4\sin^2\frac{x - y}{2}.सिद्ध कीजिए (\cos x - \cos y)^2 + (\sin x - \sin y)^2 = 4\sin^2\frac{x - y}{2}।Prove (\cos x - \cos y)^2 + (\sin x - \sin y)^2 = 4\sin^2\frac{x - y}{2}.
  1. = 2 - 2(\cos x\cos y + \sin x\sin y) = 2 - 2\cos(x - y) = 4\sin^2\frac{x - y}{2}
Q5. Prove \sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x\cos 2x\sin 4x.सिद्ध कीजिए \sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x\cos 2x\sin 4x।Prove \sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x\cos 2x\sin 4x.
  1. (\sin 7x + \sin x) + (\sin 5x + \sin 3x) = 2\sin 4x\cos 3x + 2\sin 4x\cos x
  2. = 2\sin 4x(\cos 3x + \cos x) = 2\sin 4x \cdot 2\cos 2x\cos x
Q6. Prove \frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)} = \tan 6x.सिद्ध कीजिए \frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)} = \tan 6x।Prove \frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)} = \tan 6x.
  1. \frac{2\sin 6x\cos x + 2\sin 6x\cos 3x}{2\cos 6x\cos x + 2\cos 6x\cos 3x} = \frac{\sin 6x(\cos x + \cos 3x)}{\cos 6x(\cos x + \cos 3x)} = \tan 6x
Q7. Prove \sin 3x + \sin 2x - \sin x = 4\sin x\cos\frac{x}{2}\cos\frac{3x}{2}.सिद्ध कीजिए \sin 3x + \sin 2x - \sin x = 4\sin x\cos\frac{x}{2}\cos\frac{3x}{2}।Prove \sin 3x + \sin 2x - \sin x = 4\sin x\cos\frac{x}{2}\cos\frac{3x}{2}.
  1. \sin 3x - \sin x = 2\cos 2x\sin x;\ \text{LHS} = 2\sin x\cos 2x + 2\sin x\cos x = 2\sin x(\cos 2x + \cos x)
  2. = 2\sin x \cdot 2\cos\frac{3x}{2}\cos\frac{x}{2}
Q8. \tan x = -\frac{4}{3}, x in quadrant II. Find \sin\frac{x}{2}, \cos\frac{x}{2}, \tan\frac{x}{2}.\tan x = -\frac{4}{3}, x दूसरे चतुर्थांश में। \sin\frac{x}{2}, \cos\frac{x}{2}, \tan\frac{x}{2} ज्ञात कीजिए।\tan x = -\frac{4}{3}, x in quadrant II. Find \sin\frac{x}{2}, \cos\frac{x}{2}, \tan\frac{x}{2}.
  1. \cos x = -\frac{3}{5};\ \frac{\pi}{2} < x < \pi \Rightarrow \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2} \ (\text{all positive})\cos x = -\frac{3}{5};\ \frac{\pi}{2} < x < \pi \Rightarrow \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2} \ (\text{सभी धनात्मक})\cos x = -\frac{3}{5};\ \frac{\pi}{2} < x < \pi \Rightarrow \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2} \ (\text{sab positive})
  2. \sin\frac{x}{2} = \sqrt{\frac{1 + \frac{3}{5}}{2}} = \frac{2}{\sqrt{5}},\ \cos\frac{x}{2} = \sqrt{\frac{1 - \frac{3}{5}}{2}} = \frac{1}{\sqrt{5}},\ \tan\frac{x}{2} = 2

✔ \frac{2\sqrt{5}}{5},\ \frac{\sqrt{5}}{5},\ 2

Q9. \cos x = -\frac{1}{3}, x in quadrant III. Find \sin\frac{x}{2}, \cos\frac{x}{2}, \tan\frac{x}{2}.\cos x = -\frac{1}{3}, x तीसरे चतुर्थांश में। \sin\frac{x}{2}, \cos\frac{x}{2}, \tan\frac{x}{2} ज्ञात कीजिए।\cos x = -\frac{1}{3}, x in quadrant III. Find \sin\frac{x}{2}, \cos\frac{x}{2}, \tan\frac{x}{2}.
  1. \pi < x < \frac{3\pi}{2} \Rightarrow \frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}: \ \sin > 0,\ \cos < 0
  2. \sin\frac{x}{2} = \sqrt{\frac{1 + \frac{1}{3}}{2}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3},\ \cos\frac{x}{2} = -\sqrt{\frac{1 - \frac{1}{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}
  3. \tan\frac{x}{2} = \frac{\sqrt{6}/3}{-\sqrt{3}/3} = -\sqrt{2}
Q10. \sin x = \frac{1}{4}, x in quadrant II. Find \sin\frac{x}{2}, \cos\frac{x}{2}, \tan\frac{x}{2}.\sin x = \frac{1}{4}, x दूसरे चतुर्थांश में। \sin\frac{x}{2}, \cos\frac{x}{2}, \tan\frac{x}{2} ज्ञात कीजिए।\sin x = \frac{1}{4}, x in quadrant II. Find \sin\frac{x}{2}, \cos\frac{x}{2}, \tan\frac{x}{2}.
  1. \cos x = -\frac{\sqrt{15}}{4};\ \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2} \ (\text{all positive})\cos x = -\frac{\sqrt{15}}{4};\ \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2} \ (\text{सभी धनात्मक})\cos x = -\frac{\sqrt{15}}{4};\ \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2} \ (\text{sab positive})
  2. \sin\frac{x}{2} = \sqrt{\frac{1 + \frac{\sqrt{15}}{4}}{2}} = \sqrt{\frac{4 + \sqrt{15}}{8}} = \frac{\sqrt{8 + 2\sqrt{15}}}{4}
  3. \cos\frac{x}{2} = \sqrt{\frac{4 - \sqrt{15}}{8}} = \frac{\sqrt{8 - 2\sqrt{15}}}{4},\quad \tan\frac{x}{2} = \sqrt{\frac{4 + \sqrt{15}}{4 - \sqrt{15}}} = 4 + \sqrt{15}

🎯 Quiz — 13 questions🎯 क्विज़ — 13 प्रश्न🎯 Quiz — 13 questions

Tap an option. The answer and a short reason appear at once.किसी विकल्प पर टैप करें। उत्तर और छोटा कारण तुरंत दिखेगा।Kisi option par tap kijiye. Answer aur chhota reason turant dikhega.

1. 1 radian is approximately1 रेडियन लगभग है1 radian lagbhag hai

2. An arc of 10 cm in a circle of radius 5 cm subtends an angle of5 cm त्रिज्या के वृत्त में 10 cm का चाप कितना कोण बनाता है?5 cm radius ke circle mein 10 cm ka arc kitna angle banata hai?

3. 240° in radian is240° रेडियन में है240° radian mein hai

4. If sin x = 3/5 and x is in quadrant II, cos x =यदि sin x = 3/5 और x दूसरे चतुर्थांश में है, तो cos x =Agar sin x = 3/5 aur x second quadrant mein hai, to cos x =

5. In which quadrant are tan and cot positive but sin and cos negative?किस चतुर्थांश में tan और cot धनात्मक पर sin और cos ऋणात्मक हैं?Kis quadrant mein tan aur cot positive par sin aur cos negative hain?

6. sin 765° =

7. The period of tan x istan x का आवर्तकाल हैtan x ka period hai

8. cos(x − y) equalscos(x − y) बराबर हैcos(x − y) barabar hai

9. tan 15° =

10. 1 − 2 sin² 3x equals1 − 2 sin² 3x बराबर है1 − 2 sin² 3x barabar hai

11. sin 5x + sin 3x =

12. If cos x = −1/2 and x is in quadrant III, tan x =यदि cos x = −1/2 और x तीसरे चतुर्थांश में है, तो tan x =Agar cos x = −1/2 aur x third quadrant mein hai, to tan x =

13. The maximum value of sin x cos x issin x cos x का अधिकतम मान हैsin x cos x ki maximum value hai

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These are Simple Learning Hub's own study notes and solutions, written to follow the NCERT Class 11 Mathematics textbook (Reprint 2026-27) section by section; read them with the official NCERT chapter. Diagrams are original SLH drawings.ये Simple Learning Hub के अपने स्टडी नोट्स हैं, जो NCERT कक्षा 11 गणित की किताब (2026-27) के हर अनुभाग के साथ लिखे गए हैं; इन्हें NCERT के आधिकारिक अध्याय के साथ पढ़िए। डायग्राम SLH के अपने बनाए चित्र हैं।Ye Simple Learning Hub ke apne study notes hain, jo NCERT Class 11 Maths ki kitaab (2026-27) ke har section ke saath likhe gaye hain; inhe NCERT ke official chapter ke saath padhiye. Diagrams SLH ki apni drawings hain.