Compound Angles & Identities: JEE Main Math Guide

Trigonometric identities are the algebraic backbone of JEE Main Mathematics. They underpin Calculus (integration of trig functions), Coordinate Geometry (angle between lines), and Sequences (trigonometric series). Students who have the compound angle formulae internalised as automatic reflexes solve trig questions in seconds; students who must derive them from first principles lose crucial time under pressure. This guide gives you the complete identity framework with the JEE usage pattern for each.

Compound Angle Formulae: The Foundation Layer

sin(A±B) = sinA cosB ± cosA sinB. cos(A±B) = cosA cosB ∓ sinA sinB. tan(A±B) = (tanA ± tanB)/(1 ∓ tanA tanB). These six formulae (3 additions + 3 subtractions) are the root of every trigonometric identity. Derive them once from scratch (using the unit circle or the rotation matrix), then stop re-deriving and start using. JEE exploits them in two ways: directly (evaluate sin75° = sin(45°+30°)) and indirectly (prove that some identity holds, by factoring the left side into a compound angle form).

Double angle formulae (set B=A): sin2A = 2sinA cosA, cos2A = cos²A−sin²A = 2cos²A−1 = 1−2sin²A, tan2A = 2tanA/(1−tan²A). Triple angle: sin3A = 3sinA−4sin³A, cos3A = 4cos³A−3cosA, tan3A = (3tanA−tan³A)/(1−3tan²A). These are the most directly tested trig identities — the sin3A and cos3A forms appear in integration and in series summation. Take a free trigonometry identities mock to test your recall. For the full trig foundation, see our trigonometry guide.

Product-to-Sum and Sum-to-Product Transformations

Product-to-sum: 2sinA cosB = sin(A+B)+sin(A−B); 2cosA sinB = sin(A+B)−sin(A−B); 2cosA cosB = cos(A+B)+cos(A−B); 2sinA sinB = cos(A−B)−cos(A+B). Sum-to-product: sinC+sinD = 2sin((C+D)/2)cos((C−D)/2); sinC−sinD = 2cos((C+D)/2)sin((C−D)/2); cosC+cosD = 2cos((C+D)/2)cos((C−D)/2); cosC−cosD = −2sin((C+D)/2)sin((C−D)/2).

JEE uses product-to-sum most commonly in definite integration: ∫sinmx cosnx dx → split using product-to-sum into ½∫[sin(m+n)x + sin(m−n)x]dx — two elementary integrals. Sum-to-product is used to simplify expressions like (sin70°+sin50°)/(cos70°−cos50°) — apply sum-to-product to numerator and denominator separately, then divide. These transformation techniques save 90 seconds per JEE question compared to brute-force evaluation. For integration connections, see our integration techniques guide.

Half-Angle and Auxiliary Angle Formulae

Half-angle formulae (substitute A → A/2 in double angle): sin²(A/2) = (1−cosA)/2; cos²(A/2) = (1+cosA)/2; tan(A/2) = sinA/(1+cosA) = (1−cosA)/sinA. The Weierstrass substitution: t = tan(A/2), then sinA = 2t/(1+t²), cosA = (1−t²)/(1+t²), tanA = 2t/(1−t²). This substitution converts any rational function of sinA and cosA into a rational function of t — the standard technique for integrating 1/(a+b sinx) and 1/(a+b cosx) types. It appears as an integration question in JEE every year.

Auxiliary angle form: a sinx + b cosx = R sin(x+φ) where R = √(a²+b²) and tanφ = b/a. Equivalently: R cos(x−θ) where tanθ = a/b. This form converts range-finding questions ("find maximum of 3sinx+4cosx") into trivial problems: max = R = √(9+16) = 5. JEE tests this as "find the range of 3sinx+4cosx" (answer: [−5, 5]) or "for how many x ∈ [0,2π] does 3sinx+4cosx = 5?" (answer: exactly 1, since maximum is achieved once per period). For related trigonometric equations, see our trigonometric equations guide.

Conditional Identities in Triangles

If A+B+C = π (angles of a triangle): sinA+sinB+sinC = 4cos(A/2)cos(B/2)cos(C/2); cosA+cosB+cosC = 1+4sin(A/2)sin(B/2)sin(C/2); tanA+tanB+tanC = tanA tanB tanC; sin2A+sin2B+sin2C = 4sinA sinB sinC. These conditional identities require the condition A+B+C=π and appear in the Properties of Triangles chapter, but the manipulation uses compound angle formulae. JEE asks for their values as "evaluate the expression given A+B+C=π" questions. For complete triangle properties, see our properties of triangles guide.