Sum to Product Transformations in Trigonometry
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Direct answer
Sums collapse into products through four identities: sin C + sin D = 2 sin((C + D)/2) cos((C − D)/2); sin C − sin D = 2 cos((C + D)/2) sin((C − D)/2); cos C + cos D = 2 cos((C + D)/2) cos((C − D)/2); cos C − cos D = −2 sin((C + D)/2) sin((C − D)/2). Reading them right to left gives the product-to-sum forms, e.g. 2 sin A cos B = sin(A + B) + sin(A − B). They convert sums that cannot be evaluated term-by-term into single products: sin 75° + sin 15° = 2 sin 45° cos 30° = √6/2. In triangle questions, pairing them with A + B + C = π manufactures the conditional identities — sin A + sin B + sin C = 4 cos(A/2) cos(B/2) cos(C/2) — that JEE Advanced repeatedly mines.
What you must remember
- The four sum-to-product formulas: as listed above; note the minus sign lives only in the cos C − cos D form — the detail every option set probes.
- Product-to-sum forms: 2 sin A cos B = sin(A+B) + sin(A−B); 2 cos A cos B = cos(A+B) + cos(A−B); 2 sin A sin B = cos(A−B) − cos(A+B).
- Triangle identities from sums: in any triangle, sin A + sin B + sin C = 4 cos(A/2)cos(B/2)cos(C/2) and cos A + cos B + cos C = 1 + r/R (r inradius, R circumradius) — both quotable.
- Half-angle structure in triangles: sin A + sin B = 2 cos(C/2)cos((A−B)/2), since (A + B)/2 = (π − C)/2 — the bridge to triangle geometry.
- Value gems: sin 75° + sin 15° = √6/2; sin 75° − sin 15° = √2/2; cos 15° − cos 75° = √2/2; cos 75° + cos 15° = √6/2 — four facts worth knowing cold.
- Range application: expressions like sin x + sin 5x rewritten as 2 sin 3x cos 2x expose amplitude and periodicity instantly — the method behind max/min questions.
- Equation solving: sin 3x + sin x = 0 becomes 2 sin 2x cos x = 0, splitting a sum equation into factor-solvable pieces — the most common exam use.
Where these identities earn marks
Solve sin 5x − sin 3x = sin x for x in (0, π). Convert the left side: sin 5x − sin 3x = 2 cos 4x sin x. The equation is 2 cos 4x sin x = sin x, so sin x(2 cos 4x − 1) = 0. Either sin x = 0 (rejected in the open interval) or cos 4x = 1/2, giving 4x = ±π/3 + 2kπ, so x = π/12, 5π/12, 7π/12, 11π/12 within (0, π). Direct expansion of sin 5x in powers of sin x would work eventually; the sum-to-product route takes three lines.
The triangle version shows the same economy. Prove sin A + sin B + sin C = 4 cos(A/2)cos(B/2)cos(C/2): pair sin A + sin B = 2 cos(C/2)cos((A−B)/2), then add sin C = 2 sin(C/2)cos(C/2). Factor out 2 cos(C/2): the bracket cos((A−B)/2) + sin(C/2) = cos((A−B)/2) + cos((A+B)/2) = 2 cos(A/2)cos(B/2) — where sin(C/2) = cos((π − C)/2) did the quiet work. The result: 4 cos(A/2)cos(B/2)cos(C/2), and the bound sin A + sin B + sin C ≤ 3√3/2 follows from Jensen-type inequalities — a standard Advanced companion fact.
Small identities, big saves
JEE Main plants these formulas inside 30-second simplification questions where the minus-sign placement in cos C − cos D decides between two adjacent options; memorise its sign pattern cold. Advanced builds conditional identities on triangles — given cos B, find (sin A + sin C)/sin B — where the (A + B)/2 = π/2 − C/2 conversion is used two or three times in one derivation. The recurring trap: sin x + cos x does not respond to these identities at all — it needs the auxiliary angle √2 sin(x + π/4) form. Second trap: dropping the factor 2 — the identities double the product, and a missing 2 propagates into wrong amplitude answers. Write the four formulas at the top of rough work; under pressure, half-recalled identities are more dangerous than none.
Frequently asked questions
What is sin C + sin D as a product?
2 sin((C + D)/2)·cos((C − D)/2) — the average angle in the sine, the half-difference in the cosine.
Which sum-to-product formula carries a minus sign?
cos C − cos D = −2 sin((C + D)/2)·sin((C − D)/2); the other three forms are positive-signed products.
How do you evaluate sin 75° + sin 15° without tables?
2 sin 45° cos 30° = 2·(√2/2)·(√3/2) = √6/2.
How do these identities help solve trigonometric equations?
They factor sums: sin 5x − sin 3x = sin x becomes 2 cos 4x sin x = sin x, reducing to sin x(2 cos 4x − 1) = 0 and immediate roots.
What is the triangle identity for sin A + sin B + sin C?
4 cos(A/2)·cos(B/2)·cos(C/2), derived by pairing two terms and using (A + B)/2 = π/2 − C/2.