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JEE Mains Maths · Trigonometric Identities

Sum-Product Formulas and Telescoping

Products of sines and cosines that collapse to one value, and sums that turn into products or telescope.

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Why this matters

Fourteen PYQs, eleven of them multiple choice, and three from 2026. Four are products of cosines whose angles double, or products of sines that pair up into one; five use sin θ sin(60° − θ) sin(60° + θ) = ¼ sin 3θ, its cosine twin or the factor 4cos²θ − 1; five turn a sum into a product, or split each term into a difference so the sum telescopes. Three ideas cover the page.

Concept 1 of 3: Cosine products with doubling angles

Multiply cos⁡θcos⁡2θcos⁡4θ⋯\cos\theta\cos2\theta\cos4\theta\cdots by sin⁡θ\sin\theta. Then 2sin⁡θcos⁡θ=sin⁡2θ2\sin\theta\cos\theta=\sin2\theta, which pairs with cos⁡2θ\cos2\theta, and so on. Each step doubles the angle and halves the factor, so after nn cosines only sin⁡2nθ2nsin⁡θ\frac{\sin2^n\theta}{2^n\sin\theta} is left. Products of sines often become such a chain once sin⁡x=cos⁡(90∘−x)\sin x=\cos(90^\circ-x) is used.

Definition

  • cos⁡θcos⁡2θcos⁡4θ⋯cos⁡2n−1θ=sin⁡2nθ2nsin⁡θ\cos\theta\cos2\theta\cos4\theta\cdots\cos2^{n-1}\theta=\frac{\sin2^n\theta}{2^n\sin\theta}.
  • If 2nθ=π−θ2^n\theta=\pi-\theta, the product is 12n\frac{1}{2^n}; if 2nθ=π+θ2^n\theta=\pi+\theta, it is −12n-\frac{1}{2^n}.
  • sin⁡(π−x)=sin⁡x\sin(\pi-x)=\sin x pairs equal sines; sin⁡x=cos⁡(π2−x)\sin x=\cos\left(\frac{\pi}{2}-x\right) turns sines into cosines.
  • cos⁡(π−x)=−cos⁡x\cos(\pi-x)=-\cos x replaces an angle past π2\frac{\pi}{2} by a smaller one.

Doubling product

cos⁡θcos⁡2θcos⁡4θ⋯cos⁡2n−1θ=sin⁡2nθ2nsin⁡θ\cos\theta\cos2\theta\cos4\theta\cdots\cos2^{n-1}\theta=\frac{\sin2^n\theta}{2^n\sin\theta}

Worked example

Find cos⁡π9cos⁡2π9cos⁡4π9\cos\frac{\pi}{9}\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 10 April 2023 · Q68Moderate

Example 1 · Trigonometric Identities · Sum-Product Formulas and Telescoping

 96cos⁡π33cos⁡2π33cos⁡4π33cos⁡8π33cos⁡16π33\ 96\cos\frac{\pi}{33}\cos\frac{2\pi}{33}\cos\frac{4\pi}{33}\cos\frac{8\pi}{33}\cos\frac{16\pi}{33} equal to

Check where the last angle lands

The product is ±12n\pm\frac{1}{2^n} only when sin⁡2nθ=±sin⁡θ\sin2^n\theta=\pm\sin\theta, and the sign matters. At θ=π7\theta=\frac{\pi}{7}, 8θ=π+θ8\theta=\pi+\theta, so the three-factor product is −18-\frac18, not 18\frac18.
Drill 3 more on cosine products with doubling angles

Concept 2 of 3: The 60° product identities

Expanding sin⁡(60∘−θ)sin⁡(60∘+θ)\sin(60^\circ-\theta)\sin(60^\circ+\theta) gives sin⁡260∘−sin⁡2θ=34−sin⁡2θ\sin^260^\circ-\sin^2\theta=\frac34-\sin^2\theta. Multiplied by sin⁡θ\sin\theta, that is 14sin⁡3θ\frac14\sin3\theta. So three sines at θ\theta, 60∘−θ60^\circ-\theta and 60∘+θ60^\circ+\theta make a quarter of one sine at 3θ3\theta; cosines and tangents work the same way.

Definition

  • sin⁡θsin⁡(60∘−θ)sin⁡(60∘+θ)=14sin⁡3θ\sin\theta\sin(60^\circ-\theta)\sin(60^\circ+\theta)=\frac14\sin3\theta.
  • cos⁡θcos⁡(60∘−θ)cos⁡(60∘+θ)=14cos⁡3θ\cos\theta\cos(60^\circ-\theta)\cos(60^\circ+\theta)=\frac14\cos3\theta.
  • tan⁡θtan⁡(60∘−θ)tan⁡(60∘+θ)=tan⁡3θ\tan\theta\tan(60^\circ-\theta)\tan(60^\circ+\theta)=\tan3\theta.
  • 4cos⁡2θ−1=3−4sin⁡2θ=sin⁡3θsin⁡θ4\cos^2\theta-1=3-4\sin^2\theta=\frac{\sin3\theta}{\sin\theta}, so a chain of such factors at θ,3θ,9θ,…\theta,3\theta,9\theta,\ldots telescopes.

The 60° identity

sin⁡θsin⁡(60∘−θ)sin⁡(60∘+θ)=14sin⁡3θ\sin\theta\sin(60^\circ-\theta)\sin(60^\circ+\theta)=\tfrac14\sin3\theta

Worked example

Find cos⁡10∘cos⁡50∘cos⁡70∘\cos10^\circ\cos50^\circ\cos70^\circ.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 1 · Q61Moderate

Example 2 · Trigonometric Identities · Sum-Product Formulas and Telescoping

If sin⁡(π18)sin⁡(5π18)sin⁡(7π18)=K\sin\left( \frac{\pi}{18} \right)\sin\left( \frac{5\pi}{18} \right)\sin\left( \frac{7\pi}{18} \right)= K, then the value of sin⁡(10Kπ3)\sin\left( \frac{10K\pi}{3} \right) is :

The angles must fit the pattern

sin⁡10∘sin⁡50∘sin⁡70∘\sin10^\circ\sin50^\circ\sin70^\circ fits with θ=10∘\theta=10^\circ. sin⁡10∘sin⁡20∘sin⁡40∘\sin10^\circ\sin20^\circ\sin40^\circ does not: no θ\theta gives those three angles, so the identity does not apply to it directly.
Drill 4 more on the 60° product identities

Concept 3 of 3: Sum-to-product and telescoping

The sum-to-product formulas turn a sum of two sines or cosines into one product, which often contains a known value such as cos⁡60∘\cos60^\circ. For a long sum, write each term as a difference f(k+1)−f(k)f(k+1)-f(k): the middle terms cancel and only the two ends remain.

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}.
  • For cosines at angles spaced by dd, multiply by 2sin⁡d22\sin\frac d2: 2sin⁡d2cos⁡x=sin⁡(x+d2)−sin⁡(x−d2)2\sin\frac d2\cos x=\sin\left(x+\frac d2\right)-\sin\left(x-\frac d2\right).
  • sin⁡(B−A)sin⁡Asin⁡B=cot⁡A−cot⁡B\frac{\sin(B-A)}{\sin A\sin B}=\cot A-\cot B and tan⁡θ+cot⁡θ=2sin⁡2θ\tan\theta+\cot\theta=\frac{2}{\sin2\theta}.

Sum to product

sin⁡C+sin⁡D=2sin⁡C+D2cos⁡C−D2,cos⁡C+cos⁡D=2cos⁡C+D2cos⁡C−D2\sin C+\sin D=2\sin\frac{C+D}{2}\cos\frac{C-D}{2},\quad\cos C+\cos D=2\cos\frac{C+D}{2}\cos\frac{C-D}{2}

Worked example

Find cos⁡π5+cos⁡3π5\cos\frac{\pi}{5}+\cos\frac{3\pi}{5}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 4 Apr 2026 Shift 1 · Q73Moderate

Example 3 · Trigonometric Identities · Sum-Product Formulas and Telescoping

If A=sin⁡3∘cos⁡9∘+sin⁡9∘cos⁡27∘+sin⁡27∘cos⁡81∘A = \frac{\sin 3^{\circ}}{\cos 9^{\circ}} + \frac{\sin 9^{\circ}}{\cos 27^{\circ}} + \frac{\sin 27^{\circ}}{\cos 81^{\circ}} and B=tan⁡81∘−tan⁡3∘B = \tan81^{\circ} - \tan3^{\circ}, then BA\frac{B}{A} is equal to ____\_\_\_\_ .

Multiply by the right sine

For cosines at angles spaced by dd, multiply by 2sin⁡d22\sin\frac d2, not 2sin⁡d2\sin d. With the wrong factor the new terms do not cancel in pairs.
Drill 4 more on sum-to-product and telescoping

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 (3)

  • Cosine products with doubling angles

    Doubling product

    cos⁡θcos⁡2θcos⁡4θ⋯cos⁡2n−1θ=sin⁡2nθ2nsin⁡θ\cos\theta\cos2\theta\cos4\theta\cdots\cos2^{n-1}\theta=\frac{\sin2^n\theta}{2^n\sin\theta}
  • The 60° product identities

    The 60° identity

    sin⁡θsin⁡(60∘−θ)sin⁡(60∘+θ)=14sin⁡3θ\sin\theta\sin(60^\circ-\theta)\sin(60^\circ+\theta)=\tfrac14\sin3\theta
  • Sum-to-product and telescoping

    Sum to product

    sin⁡C+sin⁡D=2sin⁡C+D2cos⁡C−D2,cos⁡C+cos⁡D=2cos⁡C+D2cos⁡C−D2\sin C+\sin D=2\sin\frac{C+D}{2}\cos\frac{C-D}{2},\quad\cos C+\cos D=2\cos\frac{C+D}{2}\cos\frac{C-D}{2}

Watch out for (3)

Test yourself on Trigonometric Identities

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