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MHT-CET Maths · Trigonometry - II

Sum-to-Product and Product Formulas

Turning a sum of two sines or cosines into a product, and a product into a sum — the moves behind ratio conditions, expressions that collapse to 1, and identities under α + β + γ = π.

All Trigonometry - II notesMHT-CET Maths strategy

Why this matters

12 PYQs, seven HARD. Seven turn a sum into a product: a ratio condition through componendo and dividendo, sin(A + B) from sin A + sin B and cos A + cos B, an expression in 20° that collapses to 1 or 4. Five go the other way: cos²48° − sin²12° (set twice in 2024), cos(log x), and a triangle identity. Two cards.

Concept 1 of 2: Sums to Products, and Ratio Conditions

A sum of two sines is twice the sine of the average angle times the cosine of the half-difference. That is why a ratio condition like 3 sin α = 5 sin β becomes a statement about half-sums and half-differences: add and subtract the two sides (componendo and dividendo) and each side factorises.

Definition

  • sin⁡C+sin⁡D=2sin⁡C+D2cos⁡C−D2\sin C + \sin D = 2\sin\frac{C + D}{2}\cos\frac{C - D}{2}, sin⁡C−sin⁡D=2cos⁡C+D2sin⁡C−D2\sin C - \sin D = 2\cos\frac{C + D}{2}\sin\frac{C - D}{2}.
  • cos⁡C+cos⁡D=2cos⁡C+D2cos⁡C−D2\cos C + \cos D = 2\cos\frac{C + D}{2}\cos\frac{C - D}{2}, cos⁡C−cos⁡D=−2sin⁡C+D2sin⁡C−D2\cos C - \cos D = -2\sin\frac{C + D}{2}\sin\frac{C - D}{2}.
  • Ratio condition: psin⁡α=qsin⁡βp\sin\alpha = q\sin\beta gives tan⁡α+β2tan⁡α−β2=q+pq−p\dfrac{\tan\frac{\alpha + \beta}{2}}{\tan\frac{\alpha - \beta}{2}} = \dfrac{q + p}{q - p}.
  • sin⁡A+sin⁡Bcos⁡A+cos⁡B=tan⁡A+B2\dfrac{\sin A + \sin B}{\cos A + \cos B} = \tan\dfrac{A + B}{2}; then sin⁡(A+B)=2t1+t2\sin(A + B) = \dfrac{2t}{1 + t^2} with that t.
  • Expressions like 3csc⁡20∘−sec⁡20∘\sqrt3\csc 20^\circ - \sec 20^\circ: write 3=2sin⁡60∘\sqrt3 = 2\sin 60^\circ (or 2cos⁡30∘2\cos 30^\circ) so the numerator becomes a single sine.

Sum to product

sin⁡C+sin⁡D=2sin⁡C+D2cos⁡C−D2cos⁡C−cos⁡D=−2sin⁡C+D2sin⁡C−D2\sin C+\sin D=2\sin\tfrac{C+D}{2}\cos\tfrac{C-D}{2} \qquad \cos C-\cos D=-2\sin\tfrac{C+D}{2}\sin\tfrac{C-D}{2}

Worked example

If 2sin⁡α=3sin⁡β2\sin\alpha = 3\sin\beta, find tan⁡α+β2tan⁡α−β2\dfrac{\tan\frac{\alpha + \beta}{2}}{\tan\frac{\alpha - \beta}{2}}.
Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 23 April Shift I · Q111Moderate

Example 1 · Trigonometry - II · Sum-to-Product and Product Formulas

If sin⁡A+sin⁡B=x\sin A + \sin B =x and cos⁡A+cos⁡B=y\cos A + \cos B =y, then sin⁡(A+B)=\sin(A + B) =

The minus sign in cos C − cos D

cos C − cos D = −2 sin((C + D)/2) sin((C − D)/2). Without the minus, cos 20° − cos 110° comes out negative when it is positive.

Inverting the ratio in componendo and dividendo

3 sin α = 5 sin β means sin α/sin β = 5/3, so the ratio is (5 + 3)/(5 − 3) = 4, not (3 + 5)/(3 − 5) = −4. Check which sine is larger first.
Drill 6 more on sums to products, and ratio conditions

Concept 2 of 2: Products to Sums, and the Two Square-Difference Identities

The product formulas are the sum-to-product ones read backwards: 2 cos A cos B is cos(A − B) + cos(A + B). Two special cases come up most often — a difference of squares of a cosine and a sine, or of two sines. Each is a product of a sum-angle and a difference-angle ratio.

Definition

  • 2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2\sin A\cos B = \sin(A + B) + \sin(A - B), 2cos⁡Acos⁡B=cos⁡(A−B)+cos⁡(A+B)2\cos A\cos B = \cos(A - B) + \cos(A + B), 2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\sin B = \cos(A - B) - \cos(A + B).
  • cos⁡2A−sin⁡2B=cos⁡(A+B)cos⁡(A−B)\cos^2 A - \sin^2 B = \cos(A + B)\cos(A - B).
  • sin⁡2A−sin⁡2B=sin⁡(A+B)sin⁡(A−B)=cos⁡2B−cos⁡2A\sin^2 A - \sin^2 B = \sin(A + B)\sin(A - B) = \cos^2 B - \cos^2 A.
  • cos⁡248∘−sin⁡212∘=cos⁡60∘cos⁡36∘=5+18\cos^2 48^\circ - \sin^2 12^\circ = \cos 60^\circ\cos 36^\circ = \dfrac{\sqrt5 + 1}{8}.
  • If α+β+γ=π\alpha + \beta + \gamma = \pi: sin⁡2α−sin⁡2γ=sin⁡(α+γ)sin⁡(α−γ)=sin⁡βsin⁡(α−γ)\sin^2\alpha - \sin^2\gamma = \sin(\alpha + \gamma)\sin(\alpha - \gamma) = \sin\beta\sin(\alpha - \gamma).

Product to sum

2cos⁡Acos⁡B=cos⁡(A−B)+cos⁡(A+B)cos⁡2A−sin⁡2B=cos⁡(A+B)cos⁡(A−B)2\cos A\cos B=\cos(A-B)+\cos(A+B) \qquad \cos^2 A-\sin^2 B=\cos(A+B)\cos(A-B)

Worked example

Find sin⁡275∘−sin⁡215∘\sin^2 75^\circ - \sin^2 15^\circ.
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 1 · Q109Hard

Example 2 · Trigonometry - II · Sum-to-Product and Product Formulas

If sin⁡18∘=5−14\sin18^\circ=\frac{\sqrt{5}-1}{4}, then cos⁡248∘−sin⁡212∘\cos^248^\circ-\sin^212^\circ has the value

Using the sine pair for a cosine-minus-sine

cos²A − sin²B = cos(A + B) cos(A − B), but sin²A − sin²B = sin(A + B) sin(A − B). Mixing them gives cos 60° sin 36° instead of cos 60° cos 36°, and that value is printed too.

Using 18° where the value needs 36°

The question gives sin 18° = (√5 − 1)/4, but the answer needs cos 36° = 1 − 2sin²18° = (√5 + 1)/4. Quoting the given value unchanged gives (√5 − 1)/8, an option.
Drill 4 more on products to sums, and the two square-difference identities

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Sums to Products, and Ratio Conditions

    Sum to product

    sin⁡C+sin⁡D=2sin⁡C+D2cos⁡C−D2cos⁡C−cos⁡D=−2sin⁡C+D2sin⁡C−D2\sin C+\sin D=2\sin\tfrac{C+D}{2}\cos\tfrac{C-D}{2} \qquad \cos C-\cos D=-2\sin\tfrac{C+D}{2}\sin\tfrac{C-D}{2}
  • Products to Sums, and the Two Square-Difference Identities

    Product to sum

    2cos⁡Acos⁡B=cos⁡(A−B)+cos⁡(A+B)cos⁡2A−sin⁡2B=cos⁡(A+B)cos⁡(A−B)2\cos A\cos B=\cos(A-B)+\cos(A+B) \qquad \cos^2 A-\sin^2 B=\cos(A+B)\cos(A-B)

Watch out for (4)

Test yourself on Trigonometry - II

15 past MHT-CET questions from this chapter, timed at 27 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Trigonometry - II chapter testDrill all 12 questions

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