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JEE Mains Maths · Trigonometric Identities

Maximum, Minimum and Range

The greatest and least values of a trigonometric expression, found from a sin θ + b cos θ or from a quantity with a known range.

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Why this matters

Nine PYQs, all of them multiple choice, and three from 2026. Four bound a sin θ + b cos θ between −√(a² + b²) and √(a² + b²), often after a rewrite; five reduce the expression to something whose range or sign is known — sin²θ cos²θ, a single sine, the signs by quadrant — or force the only solution through a discriminant. Two ideas cover the page.

Concept 1 of 2: The bound on a sin θ + b cos θ

asin⁡θ+bcos⁡θa\sin\theta+b\cos\theta is one sine wave in disguise: it equals Rsin⁡(θ+φ)R\sin(\theta+\varphi) with R=a2+b2R=\sqrt{a^2+b^2}, so it swings between −R-R and RR. Most questions first need a rewrite into this shape: expand a compound angle, or turn sin⁡2θ\sin^2\theta, cos⁡2θ\cos^2\theta and sin⁡θcos⁡θ\sin\theta\cos\theta into ratios of 2θ2\theta.

Definition

  • asin⁡θ+bcos⁡θ=Rsin⁡(θ+φ)a\sin\theta+b\cos\theta=R\sin(\theta+\varphi), R=a2+b2R=\sqrt{a^2+b^2}, tan⁡φ=ba\tan\varphi=\frac ba.
  • c+asin⁡θ+bcos⁡θc+a\sin\theta+b\cos\theta lies in [c−R, c+R][c-R,\,c+R].
  • sin⁡2θ=1−cos⁡2θ2\sin^2\theta=\frac{1-\cos2\theta}{2}, cos⁡2θ=1+cos⁡2θ2\cos^2\theta=\frac{1+\cos2\theta}{2}, sin⁡θcos⁡θ=12sin⁡2θ\sin\theta\cos\theta=\frac12\sin2\theta.
  • If c>Rc>R, 1c+asin⁡θ+bcos⁡θ\frac{1}{c+a\sin\theta+b\cos\theta} lies in [1c+R,1c−R]\left[\frac{1}{c+R},\frac{1}{c-R}\right].
  • An increasing function such as 2u2^u keeps the order, so its extremes come from the extremes of uu.

The bound

−a2+b2≤asin⁡θ+bcos⁡θ≤a2+b2-\sqrt{a^2+b^2}\le a\sin\theta+b\cos\theta\le\sqrt{a^2+b^2}

Worked example

Find the range of 3sin⁡θ−4cos⁡θ+73\sin\theta-4\cos\theta+7, and of its reciprocal.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 2 · Q63Moderate

Example 1 · Trigonometric Identities · Maximum, Minimum and Range

The least value of (cos⁡2θ−6sin⁡θcos⁡θ+3sin⁡2θ+2)\left( \cos^{2}\theta - 6\sin\theta \cos\theta + 3\sin^{2}\theta + 2 \right) is

The bound needs one angle

a2+b2\sqrt{a^2+b^2} bounds asin⁡θ+bcos⁡θa\sin\theta+b\cos\theta only when both terms have the same angle. sin⁡θ+cos⁡2θ\sin\theta+\cos2\theta is not of this form; bounding each term on its own gives a range that is too wide.
Drill 3 more on the bound on a sin θ + b cos θ

Concept 2 of 2: Ranges through a bounded quantity

When an expression reduces to one simpler quantity — p=sin⁡2θcos⁡2θp=\sin^2\theta\cos^2\theta, a single sin⁡kθ\sin k\theta, or the sign of each ratio — its range follows from the range of that quantity. When an equation can hold only at the edge of a bound, the equality case fixes the angles.

Definition

  • p=sin⁡2θcos⁡2θ∈[0,14]p=\sin^2\theta\cos^2\theta\in\left[0,\frac14\right]; sin⁡kθ,cos⁡kθ∈[−1,1]\sin k\theta,\cos k\theta\in[-1,1].
  • A linear function of pp or of sin⁡kθ\sin k\theta takes its extremes at the ends of that interval.
  • The signs by quadrant decide the sign of an expression; a sum of signs takes only a few values.
  • A quadratic in cos⁡u\cos u with a real root needs a discriminant ≥0\ge0. If that holds only with equality, the angles are fixed.

The range of p

sin⁡2θcos⁡2θ=14sin⁡22θ∈[0,14]\sin^2\theta\cos^2\theta=\tfrac14\sin^22\theta\in\left[0,\tfrac14\right]

Worked example

Find the range of sin⁡6θ+cos⁡6θ\sin^6\theta+\cos^6\theta.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 1 · Q68Moderate

Example 2 · Trigonometric Identities · Maximum, Minimum and Range

Let α\alpha and β\beta respectively be the maximum and the minimum values of the function f(θ)=4(sin⁡4(7π2−θ)+sin⁡4(11π+θ))f(\theta) = 4\left( \sin^{4}\left( \frac{7\pi}{2}- \theta \right)+\sin^{4}(11\pi + \theta) \right) −2(sin⁡6(3π2−θ)+sin⁡6(9π−θ)),θ∈R.- 2\left( \sin^{6}\left( \frac{3\pi}{2}- \theta \right)+\sin^{6}(9\pi - \theta) \right),\theta \in R. Then α+2β\alpha + 2\beta is equal to :

Check that each end is reached

The range of 1−3p1-3p uses both ends of p∈[0,14]p\in\left[0,\frac14\right]. If the question removes some angles, for example where tan⁡2θ=1\tan^2\theta=1, an end can drop out and the interval becomes open there.
Drill 4 more on ranges through a bounded quantity

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)

  • The bound on a sin θ + b cos θ

    The bound

    −a2+b2≤asin⁡θ+bcos⁡θ≤a2+b2-\sqrt{a^2+b^2}\le a\sin\theta+b\cos\theta\le\sqrt{a^2+b^2}
  • Ranges through a bounded quantity

    The range of p

    sin⁡2θcos⁡2θ=14sin⁡22θ∈[0,14]\sin^2\theta\cos^2\theta=\tfrac14\sin^22\theta\in\left[0,\tfrac14\right]

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.

Take the Trigonometric Identities chapter testDrill all 9 questions