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Compound Angle formulas

Master Compound Angle through 34 JEE Advanced-level formulas, systematically structured with every variable spelled out. Revise concept-wise, identify the areas where you need improvement, and focus your preparation with greater precision.

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Compound Angle, every formula

34 formulas, typeset and free. Print it, or keep it open beside your practice.

Pythagorean identity
$sin^{2} θ + cos^{2} θ = 1$
Q1MCQPythagorean
If $\sin\theta=\dfrac35$ and $\theta$ is acute, then $\cos\theta$ is:
- A
  $\dfrac45$
- B
  $\dfrac35$
- C
  $\dfrac54$
- D
  $-\dfrac45$
sec-tan identity
$sec^{2} θ - tan^{2} θ = 1$
Q1MCQsec-tan
If $\tan\theta=\dfrac43$, then $\sec^{2}\theta$ is:
- A
  $\dfrac{25}{9}$
- B
  $\dfrac{16}{9}$
- C
  $\dfrac79$
- D
  $\dfrac53$
cosec-cot identity
$csc^{2} θ - cot^{2} θ = 1$
Q1MCQcosec-cot
$\csc^{2}\theta-\cot^{2}\theta$ equals:
- A
  $1$
- B
  $0$
- C
  $-1$
- D
  $\csc^{2}\theta$
sin(A+B)
$sin (A + B) = sin A cos B + cos A sin B$
Q1MCQsin(A+B)
$\sin75^{\circ}=\sin(45^{\circ}+30^{\circ})$ equals:
- A
  $\dfrac{\sqrt6+\sqrt2}{4}$
- B
  $\dfrac{\sqrt6-\sqrt2}{4}$
- C
  $\dfrac{\sqrt3+1}{2}$
- D
  $\dfrac12$
sin(A-B)
$sin (A - B) = sin A cos B - cos A sin B$
cos(A+B)
$cos (A + B) = cos A cos B - sin A sin B$
Q1MCQcos(A+B)
$\cos(A+B)$ equals:
- A
  $\cos A\cos B-\sin A\sin B$
- B
  $\cos A\cos B+\sin A\sin B$
- C
  $\sin A\cos B+\cos A\sin B$
- D
  $\cos A\cos B$
cos(A-B)
$cos (A - B) = cos A cos B + sin A sin B$
Q1MCQcos(A-B)
$\cos15^{\circ}=\cos(45^{\circ}-30^{\circ})$ equals:
- A
  $\dfrac{\sqrt6+\sqrt2}{4}$
- B
  $\dfrac{\sqrt6-\sqrt2}{4}$
- C
  $\dfrac{\sqrt3-1}{2}$
- D
  $\dfrac{\sqrt3}{2}$
tan(A+B)
$tan (A + B) = \frac{tan A + tan B}{1 - tan A tan B}$
Q1MCQtan(A+B)
$\tan(A+B)$ equals:
- A
  $\dfrac{\tan A+\tan B}{1-\tan A\tan B}$
- B
  $\dfrac{\tan A-\tan B}{1+\tan A\tan B}$
- C
  $\tan A+\tan B$
- D
  $\dfrac{\tan A+\tan B}{1+\tan A\tan B}$
tan(A-B)
$tan (A - B) = \frac{tan A - tan B}{1 + tan A tan B}$
cot(A+B)
$cot (A + B) = \frac{cot A cot B - 1}{cot B + cot A}$
Q1MCQcot(A+B)
$\cot(A+B)$ equals:
- A
  $\dfrac{\cot A\cot B-1}{\cot A+\cot B}$
- B
  $\dfrac{\cot A\cot B+1}{\cot A-\cot B}$
- C
  $\cot A+\cot B$
- D
  $\dfrac{1}{\tan A+\tan B}$
sin(A+B)sin(A-B)
$sin (A + B) sin (A - B) = sin^{2} A - sin^{2} B$
Q1MCQProduct to difference
$\sin(A+B)\sin(A-B)$ equals:
- A
  $\sin^{2}A-\sin^{2}B$
- B
  $\cos^{2}A-\cos^{2}B$
- C
  $\sin^{2}A+\sin^{2}B$
- D
  $\cos^{2}A-\sin^{2}B$
cos(A+B)cos(A-B)
$cos (A + B) cos (A - B) = cos^{2} A - sin^{2} B$
Q1MCQcos·cos product
$\cos(A+B)\cos(A-B)$ equals:
- A
  $\cos^{2}A-\sin^{2}B$
- B
  $\cos^{2}A+\sin^{2}B$
- C
  $\sin^{2}A-\sin^{2}B$
- D
  $\cos^{2}B-\cos^{2}A$
sin C + sin D
$sin C + sin D = 2 sin \frac{C + D}{2} cos \frac{C - D}{2}$
Q1MCQSum to product
$\sin C+\sin D$ equals:
- A
  $2\sin\dfrac{C+D}{2}\cos\dfrac{C-D}{2}$
- B
  $2\cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$
- C
  $2\sin\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$
- D
  $\sin\dfrac{C+D}{2}$
sin C - sin D
$sin C - sin D = 2 cos \frac{C + D}{2} sin \frac{C - D}{2}$
Q1MCQDifference to product
$\sin C-\sin D$ equals:
- A
  $2\cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$
- B
  $2\sin\dfrac{C+D}{2}\cos\dfrac{C-D}{2}$
- C
  $-2\sin\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$
- D
  $2\cos\dfrac{C-D}{2}$
cos C + cos D
$cos C + cos D = 2 cos \frac{C + D}{2} cos \frac{C - D}{2}$
Q1MCQcos sum
$\cos C+\cos D$ equals:
- A
  $2\cos\dfrac{C+D}{2}\cos\dfrac{C-D}{2}$
- B
  $-2\sin\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$
- C
  $2\sin\dfrac{C+D}{2}\cos\dfrac{C-D}{2}$
- D
  $2\cos\dfrac{C-D}{2}$
cos C - cos D
$cos C - cos D = - 2 sin \frac{C + D}{2} sin \frac{C - D}{2}$
Q1MCQcos difference
$\cos C-\cos D$ equals:
- A
  $-2\sin\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$
- B
  $2\cos\dfrac{C+D}{2}\cos\dfrac{C-D}{2}$
- C
  $2\sin\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$
- D
  $-2\cos\dfrac{C+D}{2}$
2 sinA cosB
$2 sin A cos B = sin (A + B) + sin (A - B)$
Q1MCQProduct to sum
$2\sin A\cos B$ equals:
- A
  $\sin(A+B)+\sin(A-B)$
- B
  $\cos(A+B)+\cos(A-B)$
- C
  $\sin(A+B)-\sin(A-B)$
- D
  $2\sin(A+B)$
2 cosA cosB
$2 cos A cos B = cos (A + B) + cos (A - B)$
Q1MCQcos·cos to sum
$2\cos A\cos B$ equals:
- A
  $\cos(A+B)+\cos(A-B)$
- B
  $\cos(A-B)-\cos(A+B)$
- C
  $\sin(A+B)+\sin(A-B)$
- D
  $2\cos(A+B)$
tan(45°+θ)
$tan (4 5^{\circ} + θ) = \frac{1 + tan θ}{1 - tan θ} = \frac{cos θ + sin θ}{cos θ - sin θ}$
Q1MCQtan(45°+θ)
$\tan(45^{\circ}+\theta)$ equals:
- A
  $\dfrac{1+\tan\theta}{1-\tan\theta}$
- B
  $\dfrac{1-\tan\theta}{1+\tan\theta}$
- C
  $1+\tan\theta$
- D
  $\tan\theta$
sin 15° / cos 75°
$sin 1 5^{\circ} = cos 7 5^{\circ} = \frac{3 - 1}{2 2}$
Q1MCQsin 15°
$\sin15^{\circ}$ equals:
- A
  $\dfrac{\sqrt3-1}{2\sqrt2}$
- B
  $\dfrac{\sqrt3+1}{2\sqrt2}$
- C
  $\dfrac{\sqrt2-1}{2}$
- D
  $2-\sqrt3$
cos 15° / sin 75°
$cos 1 5^{\circ} = sin 7 5^{\circ} = \frac{3 + 1}{2 2}$
Q1MCQcos 15°
$\cos15^{\circ}$ equals:
- A
  $\dfrac{\sqrt3+1}{2\sqrt2}$
- B
  $\dfrac{\sqrt3-1}{2\sqrt2}$
- C
  $\dfrac{\sqrt3}{2}$
- D
  $2+\sqrt3$
tan 15°
$tan 1 5^{\circ} = 2 - 3 = cot 7 5^{\circ}$
Q1MCQtan 15°
$\tan15^{\circ}$ equals:
- A
  $2-\sqrt3$
- B
  $2+\sqrt3$
- C
  $\sqrt3-1$
- D
  $\dfrac{1}{\sqrt3}$
A+B+C=π ⇒ Σtan
$A + B + C = π \Rightarrow tan A + tan B + tan C = tan A tan B tan C$
(1+tanA)(1+tanB) when A+B=45°
$A + B = 4 5^{\circ} \Rightarrow (1 + tan A) (1 + tan B) = 2$
Three-angle tan sum
$tan (A + B + C) = \frac{S _{1} - S _{3}}{1 - S _{2}}$
S_k = k-th elementary symmetric sum of tangents
Q1MCQThree-angle tan
$\tan(A+B+C)$ equals:
- A
  $\dfrac{S_1-S_3}{1-S_2}$
- B
  $\dfrac{S_1+S_3}{1+S_2}$
- C
  $\dfrac{S_1-S_2}{1-S_3}$
- D
  $S_1-S_3$
a cosθ + b sinθ range
$- a^{2} + b^{2} \leq a cos θ + b sin θ \leq a^{2} + b^{2}$
sin^4+cos^4
$sin^{4} θ + cos^{4} θ = 1 - 2 sin^{2} θ cos^{2} θ = 1 - \frac{1}{2} sin^{2} 2 θ$
Q1MCQsin^4+cos^4
$\sin^{4}\theta+\cos^{4}\theta$ equals:
- A
  $1-\dfrac12\sin^{2}2\theta$
- B
  $1-\sin^{2}2\theta$
- C
  $1-2\sin^{2}2\theta$
- D
  $1+\dfrac12\sin^{2}2\theta$
sin^6+cos^6
$sin^{6} θ + cos^{6} θ = 1 - 3 sin^{2} θ cos^{2} θ = 1 - \frac{3}{4} sin^{2} 2 θ$
Q1MCQsin^6+cos^6
$\sin^{6}\theta+\cos^{6}\theta$ equals:
- A
  $1-\dfrac34\sin^{2}2\theta$
- B
  $1-\dfrac12\sin^{2}2\theta$
- C
  $1-3\sin^{2}2\theta$
- D
  $1-\dfrac14\sin^{2}2\theta$
cotA + tanA
$cot A + tan A = 2 csc 2 A$
Q1MCQcotA+tanA
$\cot A+\tan A$ equals:
- A
  $2\csc 2A$
- B
  $2\cot 2A$
- C
  $2\sec 2A$
- D
  $\csc A$
cotA - tanA
$cot A - tan A = 2 cot 2 A$
Q1MCQcotA-tanA
$\cot A-\tan A$ equals:
- A
  $2\cot 2A$
- B
  $2\csc 2A$
- C
  $2\tan 2A$
- D
  $\cot A$
√(1+sin 2A)
$1 + sin 2 A = ∣ cos A + sin A ∣$
Q1MCQ√(1+sin 2A)
For $A\in(0^{\circ},45^{\circ})$, $\sqrt{1+\sin 2A}$ equals:
- A
  $\cos A+\sin A$
- B
  $\cos A-\sin A$
- C
  $\sin A-\cos A$
- D
  $1+\sin A$
Period of a sin+b cos
$period of a cos (k x) + b sin (k x) = \frac{2 π}{∣ k ∣}$
Q1MCQPeriod
The period of $\sin 2x$ is:
- A
  $\pi$
- B
  $2\pi$
- C
  $\dfrac{\pi}{2}$
- D
  $4\pi$
A+B=60° result
$A + B = 6 0^{\circ} \Rightarrow sin^{2} A + sin^{2} B + sin A sin B = \frac{3}{4}$
Q1MCQA+B=60°
If $A+B=60^{\circ}$, then $\sin^{2}A+\sin^{2}B+\sin A\sin B$ equals:
- A
  $\dfrac34$
- B
  $\dfrac12$
- C
  $1$
- D
  $\dfrac{\sqrt3}{2}$
Nested radical
$2 + 2 cos θ = 2 cos \frac{θ}{2}, 0 < θ < π$
Q1MCQNested radical
The value of $\sqrt{2+2\cos\dfrac{\pi}{3}}$ is:
- A
  $\sqrt3$
- B
  $2$
- C
  $1$
- D
  $\sqrt2$

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Compound Angle formulas

Compound Angle, every formula

Pythagorean identity

sec-tan identity

cosec-cot identity

sin(A+B)

sin(A-B)

cos(A+B)

cos(A-B)

tan(A+B)

tan(A-B)

cot(A+B)

sin(A+B)sin(A-B)

cos(A+B)cos(A-B)

sin C + sin D

sin C - sin D

cos C + cos D

cos C - cos D

2 sinA cosB

2 cosA cosB

tan(45°+θ)

sin 15° / cos 75°

cos 15° / sin 75°

tan 15°

A+B+C=π ⇒ Σtan

(1+tanA)(1+tanB) when A+B=45°

Three-angle tan sum

a cosθ + b sinθ range

sin^4+cos^4

sin^6+cos^6

cotA + tanA

cotA - tanA

√(1+sin 2A)

Period of a sin+b cos

A+B=60° result

Nested radical

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