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JEE Mains Maths · Trigonometric Identities

Standard Values and Multiple Angles

Exact values at 15°, 18° and 36°, and the double- and triple-angle formulas that turn one ratio into another.

All Trigonometric Identities notesJEE Mains Maths strategy

Why this matters

Ten PYQs, eight of them multiple choice, and two from 2026. Five need an exact value at 15°, 18° or 36°, two of them to name the equation that value is a root of; five use double- or triple-angle formulas, or join cos θ and sin θ terms into one ratio. Two ideas cover the page.

Concept 1 of 2: Exact values at 15°, 18° and 36°

Beyond 30∘30^\circ, 45∘45^\circ and 60∘60^\circ, a few angles have exact surd values. 15∘15^\circ is 45∘−30∘45^\circ-30^\circ. For θ=18∘\theta=18^\circ, 2θ=90∘−3θ2\theta=90^\circ-3\theta, so sin⁡2θ=cos⁡3θ\sin2\theta=\cos3\theta; dividing by cos⁡θ\cos\theta gives 4sin⁡2θ+2sin⁡θ−1=04\sin^2\theta+2\sin\theta-1=0. The 36∘36^\circ values follow from cos⁡36∘=1−2sin⁡218∘\cos36^\circ=1-2\sin^218^\circ.

Definition

  • sin⁡15∘=6−24\sin15^\circ=\frac{\sqrt6-\sqrt2}{4}, cos⁡15∘=6+24\cos15^\circ=\frac{\sqrt6+\sqrt2}{4}, tan⁡15∘=2−3\tan15^\circ=2-\sqrt3, tan⁡75∘=2+3\tan75^\circ=2+\sqrt3.
  • sin⁡18∘=cos⁡72∘=5−14\sin18^\circ=\cos72^\circ=\frac{\sqrt5-1}{4}.
  • cos⁡36∘=sin⁡54∘=5+14\cos36^\circ=\sin54^\circ=\frac{\sqrt5+1}{4}.
  • cos⁡36∘−sin⁡18∘=12\cos36^\circ-\sin18^\circ=\frac12 and cos⁡36∘sin⁡18∘=14\cos36^\circ\sin18^\circ=\frac14.
  • To find an equation with a surd root, isolate the surd and square.

Values at 18° and 36°

sin⁡18∘=5−14,cos⁡36∘=5+14\sin18^\circ=\frac{\sqrt5-1}{4},\qquad\cos36^\circ=\frac{\sqrt5+1}{4}

Worked example

Find cos⁡236∘+sin⁡218∘\cos^236^\circ+\sin^218^\circ.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 1 · Q74Moderate

Example 1 · Trigonometric Identities · Standard Values and Multiple Angles

If cos⁡248∘−sin⁡212∘sin⁡224∘−sin⁡26∘=α+β52\frac{\cos^{2}48^{\circ} - \sin^{2}12^{\circ}}{\sin^{2}24^{\circ} - \sin^{2}6^{\circ}} = \frac{\alpha + \beta\sqrt{5}}{2}, where α,β∈N\alpha,\beta \in \mathbb{N}, then α+β\alpha + \beta is equal to ____\_\_\_\_ .

sin 18° and cos 36° differ in one sign

sin⁡18∘=5−14≈0.31\sin18^\circ=\frac{\sqrt5-1}{4}\approx0.31 and cos⁡36∘=5+14≈0.81\cos36^\circ=\frac{\sqrt5+1}{4}\approx0.81. Check the size before substituting: a value above 12\frac12 cannot be sin⁡18∘\sin18^\circ.
Drill 4 more on exact values at 15°, 18° and 36°

Concept 2 of 2: Double and triple angles

Double- and triple-angle formulas trade powers of one angle for a single ratio of a larger angle, or the reverse. Squaring sin⁡θ±cos⁡θ\sin\theta\pm\cos\theta gives sin⁡2θ\sin2\theta at once. A pair such as cos⁡θ−3sin⁡θ\cos\theta-\sqrt3\sin\theta is 2cos⁡(θ+60∘)2\cos(\theta+60^\circ), so two ratios can become one.

Definition

  • sin⁡2θ=2sin⁡θcos⁡θ\sin2\theta=2\sin\theta\cos\theta; cos⁡2θ=cos⁡2θ−sin⁡2θ=1−2sin⁡2θ=2cos⁡2θ−1\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta=2\cos^2\theta-1.
  • sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin3\theta=3\sin\theta-4\sin^3\theta; cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos3\theta=4\cos^3\theta-3\cos\theta.
  • cos⁡2θ=1−tan⁡2θ1+tan⁡2θ\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}, sin⁡2θ=2tan⁡θ1+tan⁡2θ\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}.
  • (sin⁡θ±cos⁡θ)2=1±sin⁡2θ(\sin\theta\pm\cos\theta)^2=1\pm\sin2\theta.
  • cos⁡θ+3sin⁡θ=2cos⁡(θ−60∘)\cos\theta+\sqrt3\sin\theta=2\cos(\theta-60^\circ); 3cos⁡θ+sin⁡θ=2cos⁡(θ−30∘)\sqrt3\cos\theta+\sin\theta=2\cos(\theta-30^\circ).

Double and triple angles

cos⁡2θ=1−2sin⁡2θ,sin⁡3θ=3sin⁡θ−4sin⁡3θ,cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos2\theta=1-2\sin^2\theta,\quad\sin3\theta=3\sin\theta-4\sin^3\theta,\quad\cos3\theta=4\cos^3\theta-3\cos\theta

Worked example

sin⁡θ−cos⁡θ=15\sin\theta-\cos\theta=\frac15. Find sin⁡2θ\sin2\theta and cos⁡4θ\cos4\theta.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 1 · Q68Moderate

Example 2 · Trigonometric Identities · Standard Values and Multiple Angles

The value of csc⁡10∘−3sec⁡10∘\csc10^{\circ}-\sqrt{3}\sec10^{\circ} is equal to:

sin 3θ and cos 3θ have opposite sign patterns

sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin3\theta=3\sin\theta-4\sin^3\theta, but cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos3\theta=4\cos^3\theta-3\cos\theta. Writing 3cos⁡θ−4cos⁡3θ3\cos\theta-4\cos^3\theta gives −cos⁡3θ-\cos3\theta.
Drill 4 more on double and triple angles

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)

  • Exact values at 15°, 18° and 36°

    Values at 18° and 36°

    sin⁡18∘=5−14,cos⁡36∘=5+14\sin18^\circ=\frac{\sqrt5-1}{4},\qquad\cos36^\circ=\frac{\sqrt5+1}{4}
  • Double and triple angles

    Double and triple angles

    cos⁡2θ=1−2sin⁡2θ,sin⁡3θ=3sin⁡θ−4sin⁡3θ,cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos2\theta=1-2\sin^2\theta,\quad\sin3\theta=3\sin\theta-4\sin^3\theta,\quad\cos3\theta=4\cos^3\theta-3\cos\theta

Watch out for (2)

Test yourself on Trigonometric Identities

15 past JEE Mains questions from this chapter, timed at 36 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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