Compound Angle formulas

Every formula JEE Advanced expects you to know for Compound Angle, with the variables spelled out. Free, no sign-up. Read it before a mock, not during one.

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Formulas: 34
Subject: Mathematics
Exam: JEE Advanced
No sign-in needed: ₹0

Compound Angle, every formula

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

Pythagorean identity
$sin^{2} θ + cos^{2} θ = 1$
sec-tan identity
$sec^{2} θ - tan^{2} θ = 1$
cosec-cot identity
$csc^{2} θ - cot^{2} θ = 1$
sin(A+B)
$sin (A + B) = sin A cos B + cos A sin B$
sin(A-B)
$sin (A - B) = sin A cos B - cos A sin B$
cos(A+B)
$cos (A + B) = cos A cos B - sin A sin B$
cos(A-B)
$cos (A - B) = cos A cos B + sin A sin B$
tan(A+B)
$tan (A + B) = \frac{tan A + tan B}{1 - tan A tan B}$
tan(A-B)
$tan (A - B) = \frac{tan A - tan B}{1 + tan A tan B}$
cot(A+B)
$cot (A + B) = \frac{cot A cot B - 1}{cot B + cot A}$
sin(A+B)sin(A-B)
$sin (A + B) sin (A - B) = sin^{2} A - sin^{2} B$
cos(A+B)cos(A-B)
$cos (A + B) cos (A - B) = cos^{2} A - sin^{2} B$
sin C + sin D
$sin C + sin D = 2 sin \frac{C + D}{2} cos \frac{C - D}{2}$
sin C - sin D
$sin C - sin D = 2 cos \frac{C + D}{2} sin \frac{C - D}{2}$
cos C + cos D
$cos C + cos D = 2 cos \frac{C + D}{2} cos \frac{C - D}{2}$
cos C - cos D
$cos C - cos D = - 2 sin \frac{C + D}{2} sin \frac{C - D}{2}$
2 sinA cosB
$2 sin A cos B = sin (A + B) + sin (A - B)$
2 cosA cosB
$2 cos A cos B = cos (A + B) + cos (A - B)$
tan(45°+θ)
$tan (4 5^{\circ} + θ) = \frac{1 + tan θ}{1 - tan θ} = \frac{cos θ + sin θ}{cos θ - sin θ}$
sin 15° / cos 75°
$sin 1 5^{\circ} = cos 7 5^{\circ} = \frac{3 - 1}{2 2}$
cos 15° / sin 75°
$cos 1 5^{\circ} = sin 7 5^{\circ} = \frac{3 + 1}{2 2}$
tan 15°
$tan 1 5^{\circ} = 2 - 3 = cot 7 5^{\circ}$
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
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
$sin^{4} θ + cos^{4} θ = 1 - 2 sin^{2} θ cos^{2} θ = 1 - \frac{1}{2} sin^{2} 2 θ$
sin^6+cos^6
$sin^{6} θ + cos^{6} θ = 1 - 3 sin^{2} θ cos^{2} θ = 1 - \frac{3}{4} sin^{2} 2 θ$
cotA + tanA
$cot A + tan A = 2 csc 2 A$
cotA - tanA
$cot A - tan A = 2 cot 2 A$
√(1+sin 2A)
$1 + sin 2 A = ∣ cos A + sin A ∣$
Period of a sin+b cos
$period of a cos (k x) + b sin (k x) = \frac{2 π}{∣ k ∣}$
A+B=60° result
$A + B = 6 0^{\circ} \Rightarrow sin^{2} A + sin^{2} B + sin A sin B = \frac{3}{4}$
Nested radical
$2 + 2 cos θ = 2 cos \frac{θ}{2}, 0 < θ < π$

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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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Know the Compound Angle formulas. Now use them.

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