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JEE Mains Maths · Trigonometric Identities

Powers of Sine and Cosine

Fourth, sixth and higher powers of sine and cosine, reduced to the single quantity sin²θ cos²θ or pinned down by a given condition.

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Why this matters

Ten PYQs, all of them multiple choice, and one from 2026. Six reduce powers of sine and cosine through sin²θ cos²θ, or write tan and cot through one variable; four give a condition that fixes sin²θ or sin θ cos θ first, and then evaluate. Two ideas cover the page.

Concept 1 of 2: Reducing through sin²θ cos²θ

Because sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1, every symmetric power sum of sine and cosine depends on one number, p=sin⁡2θcos⁡2θ=14sin⁡22θp=\sin^2\theta\cos^2\theta=\frac14\sin^22\theta. An expression that looks like it depends on θ\theta often turns out constant, or depends on θ\theta only through pp. With tan⁡\tan and cot⁡\cot, put t=tan⁡θt=\tan\theta instead.

Definition

  • sin⁡4θ+cos⁡4θ=1−2p\sin^4\theta+\cos^4\theta=1-2p.
  • sin⁡6θ+cos⁡6θ=1−3p\sin^6\theta+\cos^6\theta=1-3p.
  • cos⁡4θ−sin⁡4θ=cos⁡2θ\cos^4\theta-\sin^4\theta=\cos2\theta; cos⁡8θ−sin⁡8θ=cos⁡2θ (1−2p)\cos^8\theta-\sin^8\theta=\cos2\theta\,(1-2p).
  • tan⁡θ+cot⁡θ=1sin⁡θcos⁡θ=2csc⁡2θ\tan\theta+\cot\theta=\frac{1}{\sin\theta\cos\theta}=2\csc2\theta.

Power sums

sin⁡4θ+cos⁡4θ=1−2p,sin⁡6θ+cos⁡6θ=1−3p,p=sin⁡2θcos⁡2θ\sin^4\theta+\cos^4\theta=1-2p,\quad\sin^6\theta+\cos^6\theta=1-3p,\quad p=\sin^2\theta\cos^2\theta

Worked example

Show that 2(sin⁡6θ+cos⁡6θ)−3(sin⁡4θ+cos⁡4θ)2(\sin^6\theta+\cos^6\theta)-3(\sin^4\theta+\cos^4\theta) does not depend on θ\theta, and find its value.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 2 · Q63Moderate

Example 1 · Trigonometric Identities · Powers of Sine and Cosine

Let P={θ∈[0,4π]:tan⁡2θ≠1}P=\left\{ \theta\in \lbrack 0,4\pi\rbrack:\tan^{2}\theta\neq 1 \right\} and S={a∈Z:2(cos⁡8θ−sin⁡8θ)sec⁡2θ=a2,θ∈P}S=\left\{ a\in Z:2\left( \cos^{8}\theta-\sin^{8}\theta \right)\sec2\theta=a^{2},\theta\in P \right\}. Then n(S)n(S) is :

p runs from 0 to 1/4

p=sin⁡2θcos⁡2θ=14sin⁡22θp=\sin^2\theta\cos^2\theta=\frac14\sin^22\theta, so its largest value is 14\frac14, not 1. Using the range of sin⁡22θ\sin^22\theta in place of the range of pp gives a range four times too wide.
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Concept 2 of 2: A condition that fixes sin²θ

When the question gives an equation in sin⁡θ\sin\theta and cos⁡θ\cos\theta, solve it for one simple quantity first: sin⁡2θ\sin^2\theta, sin⁡θcos⁡θ\sin\theta\cos\theta, or a link such as sin⁡θ=cos⁡2θ\sin\theta=\cos^2\theta. Everything asked afterwards is substitution. An equation asin⁡4θ+bcos⁡4θ=aba+ba\sin^4\theta+b\cos^4\theta=\frac{ab}{a+b} is a perfect square in disguise.

Definition

  • asin⁡4θ+bcos⁡4θ=aba+ba\sin^4\theta+b\cos^4\theta=\frac{ab}{a+b}: multiply the right side by (sin⁡2θ+cos⁡2θ)2(\sin^2\theta+\cos^2\theta)^2; the difference is (asin⁡2θ−bcos⁡2θ)2a+b=0\frac{(a\sin^2\theta-b\cos^2\theta)^2}{a+b}=0.
  • So sin⁡2θ=ba+b\sin^2\theta=\frac{b}{a+b} and cos⁡2θ=aa+b\cos^2\theta=\frac{a}{a+b}.
  • (sin⁡θ±cos⁡θ)2=1±2sin⁡θcos⁡θ(\sin\theta\pm\cos\theta)^2=1\pm2\sin\theta\cos\theta links a sum to a product.
  • sin⁡θ+sin⁡2θ=1\sin\theta+\sin^2\theta=1 means sin⁡θ=cos⁡2θ\sin\theta=\cos^2\theta.

The perfect-square condition

asin⁡4θ+bcos⁡4θ=aba+b ⇒ sin⁡2θ=ba+ba\sin^4\theta+b\cos^4\theta=\frac{ab}{a+b}\ \Rightarrow\ \sin^2\theta=\frac{b}{a+b}

Worked example

sin⁡4θ+3cos⁡4θ=34\sin^4\theta+3\cos^4\theta=\frac34. Find tan⁡2θ\tan^2\theta.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 4 April 2025 · Q70Moderate

Example 2 · Trigonometric Identities · Powers of Sine and Cosine

If 10sin⁡4θ+15cos⁡4θ=610\sin^{4}\theta+ 15\cos^{4}\theta= 6, then the value of 27csc⁡6θ+8sec⁡6θ16sec⁡8θ\frac{27\csc^{6}\theta+ 8\sec^{6}\theta}{16\sec^{8}\theta} is:

sin²θ takes the other coefficient

In asin⁡4θ+bcos⁡4θ=aba+ba\sin^4\theta+b\cos^4\theta=\frac{ab}{a+b}, sin⁡2θ=ba+b\sin^2\theta=\frac{b}{a+b}, with bb from the cosine term. Substitute back once to check before using it.
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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)

  • Reducing through sin²θ cos²θ

    Power sums

    sin⁡4θ+cos⁡4θ=1−2p,sin⁡6θ+cos⁡6θ=1−3p,p=sin⁡2θcos⁡2θ\sin^4\theta+\cos^4\theta=1-2p,\quad\sin^6\theta+\cos^6\theta=1-3p,\quad p=\sin^2\theta\cos^2\theta
  • A condition that fixes sin²θ

    The perfect-square condition

    asin⁡4θ+bcos⁡4θ=aba+b ⇒ sin⁡2θ=ba+ba\sin^4\theta+b\cos^4\theta=\frac{ab}{a+b}\ \Rightarrow\ \sin^2\theta=\frac{b}{a+b}

Watch out for (2)

Test yourself on Trigonometric Identities

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