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🎓 Class 11 Mathematics CBSE Theory Ch 3 — Trigonometric Functions ⏱ ~15 min
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Targeting Class 11 level in Trigonometry, with Advanced difficulty.

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End-of-Chapter Exercises

This part contains full worked solutions to the four NCERT exercises in Chapter 3, followed by a chapter summary and historical note. The exercises here are meant to be tried first with pen-and-paper alongside the textbook; click "Show Solution" only after attempting the question.

Exercise 3.1 — Angle conversions and arc length
1. Find the radian measures corresponding to: (i) 25° (ii) −47°30′ (iii) 240° (iv) 520°.
(i) \(25°\times\dfrac{\pi}{180}=\dfrac{5\pi}{36}\) rad.
(ii) \(-47°30'=-\dfrac{95}{2}°\Rightarrow -\dfrac{95\pi}{360}=-\dfrac{19\pi}{72}\) rad.
(iii) \(240°\times\dfrac{\pi}{180}=\dfrac{4\pi}{3}\) rad.
(iv) \(520°\times\dfrac{\pi}{180}=\dfrac{26\pi}{9}\) rad.
2. Find the degree measures corresponding to (use \(\pi=22/7\)): (i) 11π/16 (ii) −4 (iii) 5π/3 (iv) 7π/6.
Multiply each by \(180/\pi\).
(i) \(\dfrac{11\pi}{16}\times\dfrac{180}{\pi}=\dfrac{11\times 180}{16}=\dfrac{1980}{16}=123.75°=123°45'\).
(ii) \(-4\times\dfrac{180}{\pi}=-4\times\dfrac{180\times 7}{22}=-\dfrac{2520}{11}\approx -229°5'27''\).
(iii) \(\dfrac{5\pi}{3}\times\dfrac{180}{\pi}=300°\).
(iv) \(\dfrac{7\pi}{6}\times\dfrac{180}{\pi}=210°\).
3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
360 revolutions per 60 s = 6 revolutions per second. Each revolution = \(2\pi\) rad. Hence \(6\times 2\pi=12\pi\) rad/s.
4. Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm (use \(\pi=22/7\)).
\(\theta=l/r=22/100=0.22\) rad. Convert: \(0.22\times\dfrac{180}{\pi}=0.22\times\dfrac{180\times 7}{22}=12.6°=12°36'\).
5. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of the minor arc of the chord.
Radius \(r=20\) cm, chord 20 cm. Triangle formed by chord and two radii is equilateral, so the central angle is 60° = π/3 rad. Arc length \(l=r\theta=20\times\pi/3=\dfrac{20\pi}{3}\) cm.
6. Arcs of equal length in two circles subtend angles of 60° and 75° at their centres. Find the ratio of their radii.
\(\theta_1=\pi/3\), \(\theta_2=5\pi/12\). Equal arcs ⇒ \(r_1\theta_1=r_2\theta_2\), so \(\dfrac{r_1}{r_2}=\dfrac{\theta_2}{\theta_1}=\dfrac{5\pi/12}{\pi/3}=\dfrac{5}{4}\). Ratio \(5:4\).
7. The angle through which a pendulum of length 75 cm swings if its tip describes an arc of length: (i) 10 cm, (ii) 15 cm, (iii) 21 cm.
\(\theta=l/r\) with r = 75. (i) \(10/75=2/15\) rad. (ii) \(15/75=1/5\) rad. (iii) \(21/75=7/25\) rad.
Exercise 3.2 — Find missing trig values; periodicity
1. \(\cos x=-\dfrac{1}{2}\), \(x\) in third quadrant. Find the other five.
\(\sec x=-2\). \(\sin^2 x=1-1/4=3/4\Rightarrow \sin x=\pm\sqrt 3/2\). In Q3, sin negative, so \(\sin x=-\sqrt 3/2\), \(\operatorname{cosec}x=-2/\sqrt 3\). \(\tan x=\sin/\cos=\sqrt 3\), \(\cot x=1/\sqrt 3\).
2. \(\sin x=\dfrac{3}{5}\), \(x\) in second quadrant. Find the other five.
\(\operatorname{cosec}x=5/3\). \(\cos^2 x=1-9/25=16/25\Rightarrow \cos x=\pm 4/5\); Q2 ⇒ negative, \(\cos x=-4/5\), \(\sec x=-5/4\). \(\tan x=-3/4\), \(\cot x=-4/3\).
3. \(\cot x=\dfrac{3}{4}\), \(x\) in third quadrant.
\(\tan x=4/3\). \(\sec^2 x=1+16/9=25/9\Rightarrow \sec x=\pm 5/3\); Q3 ⇒ negative, \(\sec x=-5/3\), \(\cos x=-3/5\). \(\sin x=\tan x\cdot\cos x=(4/3)(-3/5)=-4/5\), \(\operatorname{cosec}x=-5/4\).
4. \(\sec x=\dfrac{13}{5}\), \(x\) in fourth quadrant.
\(\cos x=5/13\). \(\sin^2 x=1-25/169=144/169\Rightarrow \sin x=\pm 12/13\); Q4 ⇒ negative, \(\sin x=-12/13\), \(\operatorname{cosec}x=-13/12\). \(\tan x=-12/5\), \(\cot x=-5/12\).
5. \(\tan x=-\dfrac{5}{12}\), \(x\) in second quadrant.
\(\cot x=-12/5\). \(\sec^2 x=1+25/144=169/144\Rightarrow \sec x=\pm 13/12\); Q2 ⇒ negative, \(\sec x=-13/12\), \(\cos x=-12/13\). \(\sin x=\tan x\cdot\cos x=(-5/12)(-12/13)=5/13\), \(\operatorname{cosec}x=13/5\).
6. Find the value of \(\sin 765°\).
\(765°=720°+45°=2\times 360°+45°\). By periodicity, \(\sin 765°=\sin 45°=1/\sqrt 2\).
7. \(\operatorname{cosec}(-1410°)\).
\(-1410°=-4\times 360°+30°\). \(\sin(-1410°)=\sin 30°=1/2\), so cosec = 2.
8. \(\tan\dfrac{19\pi}{3}\).
\(\dfrac{19\pi}{3}=6\pi+\dfrac{\pi}{3}\). Since tan has period π (and 6π = 6×π), \(\tan(19\pi/3)=\tan(\pi/3)=\sqrt 3\).
9. \(\sin\!\left(-\dfrac{11\pi}{3}\right)\).
\(-\dfrac{11\pi}{3}=-4\pi+\dfrac{\pi}{3}\). So \(\sin(-11\pi/3)=\sin(\pi/3)=\sqrt 3/2\).
10. \(\cot\!\left(-\dfrac{15\pi}{4}\right)\).
\(-\dfrac{15\pi}{4}=-4\pi+\dfrac{\pi}{4}\). So \(\cot(-15\pi/4)=\cot(\pi/4)=1\).
Exercise 3.3 — Identities (selected solutions)

Exercise 3.3 contains 25 questions on identity verification. Below are full solutions to ten representative ones; the techniques generalise immediately to the rest. (Q1–Q4 use direct table values; Q5–Q10 use sum/difference; Q11–Q20 use sum-to-product; Q21–Q25 use multiple-angle.)

1. Prove: \(\sin^2\!\dfrac{\pi}{6}+\cos^2\!\dfrac{\pi}{3}-\tan^2\!\dfrac{\pi}{4}=-\dfrac{1}{2}\).
\(\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-1^2=\dfrac{1}{4}+\dfrac{1}{4}-1=-\dfrac{1}{2}\). \(\square\)
5. (i) Find \(\sin 75°\).   (ii) \(\tan 15°\).
(i) \(\sin 75°=\sin(45°+30°)=\sin 45°\cos 30°+\cos 45°\sin 30°=\dfrac{\sqrt 6+\sqrt 2}{4}=\dfrac{\sqrt 3+1}{2\sqrt 2}\).
(ii) \(\tan 15°=\tan(45°-30°)=\dfrac{1-1/\sqrt 3}{1+1/\sqrt 3}=\dfrac{\sqrt 3-1}{\sqrt 3+1}=\dfrac{(\sqrt 3-1)^2}{2}=2-\sqrt 3\).
10. Prove: \(\sin(n+1)x\sin(n+2)x+\cos(n+1)x\cos(n+2)x=\cos x\).
LHS \(=\cos\bigl[(n+2)x-(n+1)x\bigr]=\cos x\) by Identity 4. \(\square\)
12. Prove: \(\sin^2 6x-\sin^2 4x=\sin 2x\sin 10x\).
LHS = \((\sin 6x+\sin 4x)(\sin 6x-\sin 4x)\). By sum-to-product: \(\sin 6x+\sin 4x=2\sin 5x\cos x\) and \(\sin 6x-\sin 4x=2\cos 5x\sin x\). Product = \(4\sin 5x\cos 5x\sin x\cos x=(2\sin 5x\cos 5x)(2\sin x\cos x)=\sin 10x\sin 2x\). \(\square\)
17. Prove: \(\dfrac{\sin 5x+\sin 3x}{\cos 5x+\cos 3x}=\tan 4x\).
Numerator \(=2\sin 4x\cos x\); denominator \(=2\cos 4x\cos x\). Ratio \(=\tan 4x\). \(\square\)
18. Prove: \(\dfrac{\sin x-\sin y}{\cos x+\cos y}=\tan\dfrac{x-y}{2}\).
Numerator \(=2\cos\dfrac{x+y}{2}\sin\dfrac{x-y}{2}\); denominator \(=2\cos\dfrac{x+y}{2}\cos\dfrac{x-y}{2}\). Ratio simplifies (the \(\cos\dfrac{x+y}{2}\) cancels) to \(\tan\dfrac{x-y}{2}\). \(\square\)
19. Prove: \(\dfrac{\sin x+\sin 3x}{\cos x+\cos 3x}=\tan 2x\).
Numerator \(=2\sin 2x\cos x\); denominator \(=2\cos 2x\cos x\). Ratio \(=\tan 2x\). \(\square\)
22. Prove: \(\cot x\cot 2x-\cot 2x\cot 3x-\cot 3x\cot x=1\).
Use \(3x=2x+x\) so \(\cot 3x=\dfrac{\cot 2x\cot x-1}{\cot x+\cot 2x}\). Cross-multiply: \(\cot 3x(\cot x+\cot 2x)=\cot x\cot 2x-1\), i.e. \(\cot x\cot 2x-\cot 2x\cot 3x-\cot 3x\cot x=1\). \(\square\)
24. Prove: \(\cos 4x=1-8\sin^2 x\cos^2 x\).
\(\cos 4x=1-2\sin^2 2x=1-2(2\sin x\cos x)^2=1-8\sin^2 x\cos^2 x\). \(\square\)
25. Prove: \(\cos 6x=32\cos^6 x-48\cos^4 x+18\cos^2 x-1\).
\(\cos 6x=2\cos^2 3x-1\). Substitute \(\cos 3x=4\cos^3 x-3\cos x\): \((4c^3-3c)^2=16c^6-24c^4+9c^2\) (where \(c=\cos x\)). Then \(2(16c^6-24c^4+9c^2)-1=32c^6-48c^4+18c^2-1\). \(\square\)
Miscellaneous Exercise on Chapter 3
1. Prove: \(2\cos\dfrac{\pi}{13}\cos\dfrac{9\pi}{13}+\cos\dfrac{3\pi}{13}+\cos\dfrac{5\pi}{13}=0\).
\(2\cos A\cos B=\cos(A-B)+\cos(A+B)\). With \(A=\pi/13,\ B=9\pi/13\): the first term \(=\cos(8\pi/13)+\cos(10\pi/13)=-\cos(5\pi/13)-\cos(3\pi/13)\) (using \(\cos(\pi-\theta)=-\cos\theta\)). Adding the remaining two terms: \(-\cos(5\pi/13)-\cos(3\pi/13)+\cos(3\pi/13)+\cos(5\pi/13)=0\). \(\square\)
3. Prove: \((\cos x+\cos y)^2+(\sin x-\sin y)^2=4\cos^2\!\dfrac{x+y}{2}\).
\((\cos x+\cos y)^2=4\cos^2\dfrac{x+y}{2}\cos^2\dfrac{x-y}{2}\). \((\sin x-\sin y)^2=4\cos^2\dfrac{x+y}{2}\sin^2\dfrac{x-y}{2}\). Sum = \(4\cos^2\dfrac{x+y}{2}\bigl(\cos^2\dfrac{x-y}{2}+\sin^2\dfrac{x-y}{2}\bigr)=4\cos^2\dfrac{x+y}{2}\). \(\square\)
5. Prove: \(\sin x+\sin 3x+\sin 5x+\sin 7x=4\cos x\cos 2x\sin 4x\).
Group: \((\sin x+\sin 7x)+(\sin 3x+\sin 5x)=2\sin 4x\cos 3x+2\sin 4x\cos x=2\sin 4x(\cos 3x+\cos x)\). Now \(\cos 3x+\cos x=2\cos 2x\cos x\). So total \(=4\sin 4x\cos 2x\cos x\). \(\square\)
7. Prove: \(\sin 3x+\sin 2x-\sin x=4\sin x\cos\dfrac{x}{2}\cos\dfrac{3x}{2}\).
\(\sin 3x-\sin x=2\cos 2x\sin x\). Add \(\sin 2x=2\sin x\cos x\): total \(=2\sin x(\cos 2x+\cos x)=2\sin x\cdot 2\cos\dfrac{3x}{2}\cos\dfrac{x}{2}=4\sin x\cos\dfrac{x}{2}\cos\dfrac{3x}{2}\). \(\square\)
8. \(\tan x=-\dfrac{4}{3},\ x\) in Q2. Find \(\sin\dfrac{x}{2},\ \cos\dfrac{x}{2},\ \tan\dfrac{x}{2}\).
Q2 ⇒ \(\cos x<0\). \(\sec^2 x=1+16/9=25/9\Rightarrow\cos x=-3/5\).
\(x\in(\pi/2,\pi)\Rightarrow x/2\in(\pi/4,\pi/2)\Rightarrow\) Q1, both sin(x/2) and cos(x/2) positive.
\(2\sin^2(x/2)=1-\cos x=1+3/5=8/5\Rightarrow\sin(x/2)=2/\sqrt 5\).
\(2\cos^2(x/2)=1+\cos x=2/5\Rightarrow\cos(x/2)=1/\sqrt 5\).
\(\tan(x/2)=2\).
9. \(\cos x=-\dfrac{1}{3},\ x\) in Q3. Find \(\sin\dfrac{x}{2},\ \cos\dfrac{x}{2},\ \tan\dfrac{x}{2}\).
Q3 ⇒ \(x\in(\pi,3\pi/2)\Rightarrow x/2\in(\pi/2,3\pi/4)\Rightarrow\) Q2. So sin(x/2) > 0, cos(x/2) < 0.
\(2\sin^2(x/2)=1-\cos x=1+1/3=4/3\Rightarrow\sin(x/2)=\sqrt{2/3}=\sqrt 6/3\).
\(2\cos^2(x/2)=1+\cos x=2/3\Rightarrow\cos(x/2)=-\sqrt{1/3}=-\sqrt 3/3\).
\(\tan(x/2)=-\sqrt 2\).
10. \(\sin x=\dfrac{1}{4},\ x\) in Q2. Find \(\sin\dfrac{x}{2},\ \cos\dfrac{x}{2},\ \tan\dfrac{x}{2}\).
Q2 ⇒ \(\cos x<0\). \(\cos^2 x=1-1/16=15/16\Rightarrow\cos x=-\sqrt{15}/4\).
\(x/2\in(\pi/4,\pi/2)\Rightarrow\) Q1, both positive.
\(2\sin^2(x/2)=1-\cos x=1+\sqrt{15}/4=\dfrac{4+\sqrt{15}}{4}\Rightarrow\sin(x/2)=\sqrt{\dfrac{4+\sqrt{15}}{8}}\).
\(2\cos^2(x/2)=\dfrac{4-\sqrt{15}}{4}\Rightarrow\cos(x/2)=\sqrt{\dfrac{4-\sqrt{15}}{8}}\).
\(\tan(x/2)=\sqrt{\dfrac{4+\sqrt{15}}{4-\sqrt{15}}}\). Rationalising: numerator·(4+√15)/(4²−15) = (4+√15)/1 = 4+√15. So \(\tan(x/2)=4+\sqrt{15}\).
Activity: Identity Toolkit Cheat-Sheet
L3 Apply
Materials: One A4 sheet, two coloured pens.
Predict: If you had to choose just five identities to memorise that would let you derive everything else, which five would they be?
  1. On the left half of the sheet, list the five "core" identities you'd retain.
  2. On the right half, derive the following from those five only: \(\cos 2x\), \(\sin 2x\), \(\tan(x-y)\), \(\sin x+\sin y\), \(\cos(\pi/2-x)\).
  3. If your derivation requires an identity you didn't list, mark it red — that identity is also "core".
  4. Refine your "core" list and repeat for: \(\cos 3x\), \(\sin x-\sin y\), \(\tan(x+y)\), half-angle formulas.
A defensible "core" set: (1) \(\sin^2+\cos^2=1\), (2) \(\cos(x+y)=\cos x\cos y-\sin x\sin y\), (3) \(\sin(x+y)=\sin x\cos y+\cos x\sin y\), (4) the parity relations \(\sin(-x)=-\sin x,\ \cos(-x)=\cos x\), and (5) the cofunction \(\sin(\pi/2-x)=\cos x\). Every other identity in the chapter follows from these five.

Consolidation Competency-Based Questions

Scenario: A high-school student is preparing for a board examination on Chapter 3. She wants to identify which solution technique to apply for any given identity-proof question on first read.
Q1. The expression \(\dfrac{\cos 5x-\cos 3x}{\sin 5x+\sin 3x}\) simplifies to:
L3 Apply
  • (a) \(\tan x\)
  • (b) \(-\tan x\)
  • (c) \(\cot 4x\)
  • (d) \(-\cot 4x\)
Answer: (b). Numerator \(=-2\sin 4x\sin x\); denominator \(=2\sin 4x\cos x\). Ratio \(=-\tan x\).
Q2. Given the multiple-choice rule "If the question contains \(\sin A\pm\sin B\) or \(\cos A\pm\cos B\), apply sum-to-product first." Apply it to: \(\sin 75°-\sin 15°\). What does it equal?
L3 Apply
Answer: \(2\cos 45°\sin 30°=2\cdot\dfrac{1}{\sqrt 2}\cdot\dfrac{1}{2}=\dfrac{1}{\sqrt 2}\).
Q3. Analyse: which of the following identities CANNOT be proved using only the chord-distance proof of cos(x+y) (Identity 3)?
L4 Analyse
  • (a) \(\cos(x-y)\) formula
  • (b) \(\sin(x+y)\) formula
  • (c) \(\sin^2 x+\cos^2 x=1\)
  • (d) \(\tan(x+y)\) formula
Answer: (c). The Pythagorean identity is the input to the chord-distance proof, not its output. The proof uses \(\cos^2+\sin^2=1\) (which itself comes from the unit-circle definition). The other three (a, b, d) all follow from Identity 3 by chains of substitutions.
Q4. (True/False) "Knowing only \(\cos 0,\ \cos 30°,\ \cos 45°,\ \cos 60°,\ \cos 90°\) and the sum/difference identities, one can compute the cosine of any multiple of 15°." Justify.
L5 Evaluate
True. 15° = 45° − 30°, so cos 15° comes from cos(45° − 30°). Then 75° = 45° + 30°. By symmetry and supplements, every multiple of 15° in [0°, 360°] is reachable. Even cos 75° = sin 15° via the cofunction. The grid is closed under the sum/difference operations.
Q5. Design: write a short "decision tree" (3 questions) that decides which technique to use on a trig identity-proof problem. (Sum-to-product? Multiple-angle? Sum/difference? Reduction by π/2 or π?)
L6 Create
Sample decision tree:
(1) Does the expression contain a SUM of two trig terms with same function (e.g. \(\sin A+\sin B\))? → use sum-to-product.
(2) Does it contain a DOUBLE or TRIPLE angle (e.g. \(\sin 2x,\cos 3x,\tan 4x\))? → use multiple-angle identity.
(3) Does it contain an angle like \(\pi-x,\ \pi/2+x,\ 2\pi-x\)? → use the reduction formulas (Part 3 Identity 9). If none of the above, fall back to writing things in terms of \(\sin x,\cos x\) and using \(\sin^2+\cos^2=1\).

Consolidation Assertion–Reason

Assertion (A): \(\sin 765°=\dfrac{1}{\sqrt 2}\).
Reason (R): sin is periodic with period \(2\pi\) (i.e. 360°), so adding any integer multiple of 360° leaves the value unchanged.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). 765° = 720° + 45° = 2·360° + 45°. R is the principle that makes A possible.
Assertion (A): The identity \(\cot x\cot 2x-\cot 2x\cot 3x-\cot 3x\cot x=1\) holds for all valid x.
Reason (R): It follows from \(\cot 3x=\cot(2x+x)\) and the cot-sum formula \(\cot(A+B)=(\cot A\cot B-1)/(\cot A+\cot B)\).
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). Cross-multiplying R gives exactly A.
Assertion (A): \(\tan(19\pi/3)=\sqrt 3\).
Reason (R): \(\tan\) has fundamental period π.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). \(19\pi/3=6\pi+\pi/3\); since 6π is an integer multiple of π, tan(19π/3)=tan(π/3)=√3. R is the precise reason.

Chapter Summary

Key formulas at a glance
  • Arc length: \(l=r\theta\) (θ in radians).
  • Conversion: radians \(=\dfrac{\pi}{180}\times\) degrees,   degrees \(=\dfrac{180}{\pi}\times\) radians.
  • Pythagorean: \(\sin^2 x+\cos^2 x=1\),   \(1+\tan^2 x=\sec^2 x\),   \(1+\cot^2 x=\operatorname{cosec}^2 x\).
  • Periodicity: \(\cos(2n\pi+x)=\cos x\), \(\sin(2n\pi+x)=\sin x\); period of tan and cot is π.
  • Parity: \(\sin(-x)=-\sin x\) (odd), \(\cos(-x)=\cos x\) (even).
  • Cofunctions: \(\cos(\pi/2-x)=\sin x\), \(\sin(\pi/2-x)=\cos x\).
  • Sum: \(\cos(x\pm y)=\cos x\cos y\mp\sin x\sin y\),   \(\sin(x\pm y)=\sin x\cos y\pm\cos x\sin y\).
  • Tangent sum: \(\tan(x\pm y)=\dfrac{\tan x\pm\tan y}{1\mp\tan x\tan y}\).
  • Double angle: \(\sin 2x=2\sin x\cos x\); \(\cos 2x=\cos^2 x-\sin^2 x=2\cos^2 x-1=1-2\sin^2 x\); \(\tan 2x=\dfrac{2\tan x}{1-\tan^2 x}\).
  • Triple angle: \(\sin 3x=3\sin x-4\sin^3 x\); \(\cos 3x=4\cos^3 x-3\cos x\); \(\tan 3x=\dfrac{3\tan x-\tan^3 x}{1-3\tan^2 x}\).
  • Sum-to-product: \(\sin A+\sin B=2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2}\), \(\sin A-\sin B=2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2}\), \(\cos A+\cos B=2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2}\), \(\cos A-\cos B=-2\sin\dfrac{A+B}{2}\sin\dfrac{A-B}{2}\).

Historical Note

The study of trigonometry was first carried out in India. The Indian mathematicians Aryabhata (476 CE), Brahmagupta (598), Bhaskara I (600) and Bhaskara II (1114) gave important results. All this knowledge first went from India to middle East and from there to Europe. The Greeks had also studied the subject of trigonometry but their approach was so clumsy that when the Indian approach became known, it was immediately adopted throughout the world.

In India, the predecessor of the modern trigonometric functions, known as the sine of an angle, and the introduction of the sine function represents the main contribution of the siddhantas (Sanskrit astronomical works) to the history of mathematics.

Bhaskara I (about 600) gave formulae to find the values of sine functions for angles more than 90°. A sixteenth-century Malayalam work Yuktibhasa (period) contains a proof for the expansion of \(\sin(A+B)\). Exact expression for sines or cosines of 18°, 36°, 54°, 72°, etc. are given by Bhaskara II.

Bhaskara I

c. 600 – 680 CE

An Indian mathematician of the school of Aryabhata, Bhaskara I is best known for his rational approximation \(\sin x\approx\dfrac{16x(\pi-x)}{5\pi^2-4x(\pi-x)}\) (in modern notation), accurate to 4 decimal places without calculus or Taylor series. He also extended the sine function to angles greater than 90°, completing what Aryabhata had begun.

Frequently Asked Questions

What is the radian measure of 240°?
240° × π/180 = 4π/3 radians.
How many radians does a wheel turn per second if it makes 360 revolutions per minute?
360 rpm = 6 revolutions per second = 6 × 2π = 12π radians per second.
What is the value of sin 765°?
765° = 720° + 45° = 2 × 360° + 45°. By periodicity, sin 765° = sin 45° = 1/√2.
What is tan 19π/3?
19π/3 = 6π + π/3. Since tan has period π and 6π is an integer multiple of π, tan(19π/3) = tan(π/3) = √3.
What is the chapter summary of Trigonometric Functions Class 11?
Key formulas: l = rθ; π radians = 180°; the unit-circle definitions of all six trig functions; sin²x + cos²x = 1; periodicity of sin and cos with period 2π; the sum/difference formulas for sin, cos, tan, cot; the double-angle and triple-angle identities; and the sum-to-product / product-to-sum identities.
Who developed the modern sine function?
The Indian astronomer Aryabhata (476–550 CE) introduced the half-chord sine and tabulated its values. Bhaskara I (600 CE) gave a remarkable rational approximation to sin x. The modern symbol 'sin' traces back through Arabic to Aryabhata's Sanskrit term jya-ardha.
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