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SSC CGL Maths · Trigonometric Ratios and Identities

Trigonometric Equations and Maximum Values

Solve for θ by turning the equation into a single ratio and reading the table, and find how large or small an expression in sin θ and cos θ can be.

All Trigonometric Ratios and Identities notes

Why this matters

15 questions in the bank, 12 of them from 2024. The equations almost always land on 30°, 45° or 60°, so with angle options the quickest check is to put each option in. The greatest-value questions rest on one fact: a sin θ + b cos θ never passes √(a² + b²).

Concept 1 of 2: Reduce to one ratio, then read the angle

An equation with two ratios in it cannot be read off the table. Use an identity to leave just one ratio, solve for it like ordinary algebra, and the table gives the angle.

Definition

  • Mixed squares: in asin⁡2θ+bcos⁡2θ=ca\sin^2\theta + b\cos^2\theta = c, replace cos⁡2θ\cos^2\theta by 1−sin⁡2θ1 - \sin^2\theta. This gives (a−b)sin⁡2θ=c−b(a - b)\sin^2\theta = c - b.
  • cosec with cot, or sec with tan: replace cosec2θ\text{cosec}^2\theta by 1+cot⁡2θ1 + \cot^2\theta (or sec⁡2θ\sec^2\theta by 1+tan⁡2θ1 + \tan^2\theta).
  • A quadratic in one ratio: factor it, then reject any root with sin⁡θ\sin\theta or cos⁡θ\cos\theta outside −1 to 1.
  • A ratio plus its reciprocal equal to 2, such as tan⁡θ+cot⁡θ=2\tan\theta + \cot\theta = 2, forces the ratio to be 1.
  • Cancel a factor such as cos⁡θ\cos\theta only after checking it is not 0 in the given range; otherwise you may lose a solution.
  • Fast route: when the options are angles, put each one into the equation with the table values and keep the one that fits. Check the range given: 90° may be excluded.

Mixed squares of sin and cos

asin⁡2θ+bcos⁡2θ=c⇒sin⁡2θ=c−ba−ba\sin^2\theta + b\cos^2\theta = c \Rightarrow \sin^2\theta = \frac{c - b}{a - b}

Worked example

If 7sin⁡2θ+3cos⁡2θ=67\sin^2\theta + 3\cos^2\theta = 6 and θ is acute, find θ and tan⁡θ\tan\theta.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 18 Sep 2025, Shift 1 · Q72Easy

Example 1 · Trigonometric Ratios and Identities · Trigonometric Equations and Maximum Values

If sin⁡2A−cos⁡2A=12\sin^2 A - \cos^2 A = \frac{1}{2}, find the value of cos⁡2A\cos^2 A.

Keeping an impossible root

A quadratic in sin⁡θ\sin\theta often gives a root such as sin⁡θ=−2\sin\theta = -2 or 32\frac{3}{2}. No angle has a sine outside −1 to 1, so that root is thrown away, not reported.

Cancelling a factor that can be zero

From cot⁡2θ=2cos⁡2θ\cot^2\theta = 2\cos^2\theta, cancelling cos⁡2θ\cos^2\theta loses θ = 90°, where both sides are 0. Cancel only when the range rules that value out, and say so.
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Concept 2 of 2: Greatest and least values of a sin θ + b cos θ and of powers

sin θ and cos θ cannot both be large at once, so asin⁡θ+bcos⁡θa\sin\theta + b\cos\theta tops out below a + b, at a2+b2\sqrt{a^2 + b^2}. It reaches that top when sin θ and cos θ are in the same ratio as a and b.

Definition

  • −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}.
  • The top is reached when sin⁡θ=aa2+b2\sin\theta = \frac{a}{\sqrt{a^2 + b^2}} and cos⁡θ=ba2+b2\cos\theta = \frac{b}{\sqrt{a^2 + b^2}}. So if asin⁡θ+bcos⁡θa\sin\theta + b\cos\theta equals its greatest value, tan⁡θ=ab\tan\theta = \frac{a}{b}.
  • An equation asin⁡θ+bcos⁡θ=ca\sin\theta + b\cos\theta = c with ∣c∣>a2+b2|c| > \sqrt{a^2 + b^2} has no solution. With a smaller c, divide by a2+b2\sqrt{a^2 + b^2} and use a compound angle: 3cos⁡θ+sin⁡θ=2cos⁡(θ−30∘)\sqrt{3}\cos\theta + \sin\theta = 2\cos(\theta - 30^\circ).
  • Powers shrink: for 0≤x≤10 \le x \le 1, xn≤xx^n \le x. So any higher even powers satisfy sin⁡2mθ+cos⁡2nθ≤sin⁡2θ+cos⁡2θ=1\sin^{2m}\theta + \cos^{2n}\theta \le \sin^2\theta + \cos^2\theta = 1, with 1 reached at θ = 0°.
  • sin⁡4θ+cos⁡4θ=1−2sin⁡2θcos⁡2θ\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta lies from 12\frac{1}{2} (at 45°) to 1 (at 0°).

Range of a sin θ + b cos θ

−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 greatest value of 5sin⁡θ+12cos⁡θ5\sin\theta + 12\cos\theta, and tan⁡θ\tan\theta where it is reached.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 19 Sep 2024, Shift 2 · Q71Easy

Example 2 · Trigonometric Ratios and Identities · Trigonometric Equations and Maximum Values

The maximum value of (2sin⁡θ+3cos⁡θ)(2 \sin \theta + 3 \cos \theta) is:

Adding the coefficients

The greatest value of 6sin⁡θ+8cos⁡θ6\sin\theta + 8\cos\theta is not 14, because sin θ and cos θ are never both 1. It is 36+64=10\sqrt{36 + 64} = 10.

Forgetting the least value is negative

The range of asin⁡θ+bcos⁡θa\sin\theta + b\cos\theta is symmetric: if the greatest value is 13, the least is −13, not 0.
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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)

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Test yourself on Trigonometric Ratios and Identities

20 past SSC CGL questions from this chapter, timed at 12 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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