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CDS Mathematics · Trigonometric Ratios and Identities

Trigonometric Equations

Turn the equation into one ratio, solve it as an ordinary equation, then throw out every root the ratio or the stated range cannot allow.

All Trigonometric Ratios and Identities notesCDS Mathematics strategy

Why this matters

Twenty-six PYQs, mostly MODERATE, and nearly every one ends by asking for some other quantity once θ is known. The solving is routine; the marks are lost at the last step, where a root outside [−1, 1] or outside the stated interval has to be rejected.

Concept 1 of 4: Reduce to one ratio and solve the quadratic

An equation with both sin⁡2θ\sin^2\theta and cos⁡2θ\cos^2\theta, or both tan⁡2θ\tan^2\theta and sec⁡θ\sec\theta, becomes an ordinary quadratic once one ratio is written in terms of the other. Solve it, then keep only the roots a ratio can actually take and the stated range allows.

Definition

  • Replace cos⁡2θ\cos^2\theta by 1−sin⁡2θ1 - \sin^2\theta (or the reverse), or tan⁡2θ\tan^2\theta by sec⁡2θ−1\sec^2\theta - 1.
  • Solve the resulting quadratic in the single ratio.
  • Reject a root that the ratio cannot take (sin⁡θ=3\sin\theta = 3, sec⁡θ=12\sec\theta = \dfrac12), and a root outside the stated interval.
  • A 'how many solutions' question is answered by counting the roots that survive — often zero.

The substitutions

cos⁡2θ=1−sin⁡2θ,tan⁡2θ=sec⁡2θ−1\cos^2\theta = 1 - \sin^2\theta, \qquad \tan^2\theta = \sec^2\theta - 1

Worked example

Solve 2sin⁡2θ−3cos⁡θ=02\sin^2\theta - 3\cos\theta = 0 for 0∘<θ<90∘0^\circ < \theta < 90^\circ.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2023 · CDS (I) 2023 — Elementary Mathematics · Q53Moderate

Example 1 · Trigonometric Ratios and Identities · Trigonometric Equations

If 2cos⁡2θ+sin⁡θ−2=02\cos^2\theta + \sin\theta - 2 = 0, 0<θ≤π20 < \theta \le \frac{\pi}{2}, then what is the value of θ\theta ?

A strict interval can exclude the only root

If the only surviving root is sin⁡θ=1\sin\theta = 1, that is θ=90∘\theta = 90^\circ. On 0<θ<π20 < \theta < \dfrac{\pi}{2} — a strict inequality — it is excluded, and the equation has no solution. Read the inequality signs of the range before answering.
Drill 15 more on reduce to one ratio and solve the quadratic

Concept 2 of 4: Equations in a ratio and its reciprocal

Equations like sec⁡θ+cos⁡θ=52\sec\theta + \cos\theta = \dfrac52 or 12(tan⁡θ+cot⁡θ)=2512(\tan\theta + \cot\theta) = 25 are t+1t=ct + \dfrac1t = c in disguise. Multiply through by tt to get a quadratic; its two roots are reciprocals of each other, and the stated range picks one.

Definition

  • t+1t=ct + \dfrac1t = c becomes t2−ct+1=0t^2 - ct + 1 = 0, whose roots multiply to 11.
  • For cos⁡θ+sec⁡θ\cos\theta + \sec\theta, the root with ∣t∣≤1|t| \le 1 is the cosine.
  • For tan⁡θ+cot⁡θ\tan\theta + \cot\theta, the two roots are tan⁡θ\tan\theta for two complementary angles; a range like 45∘<θ<90∘45^\circ < \theta < 90^\circ picks the root above 11.
  • t+1t=2t + \dfrac1t = 2 forces t=1t = 1: the sum of a positive number and its reciprocal is 22 only at 11.

Clearing the reciprocal

t+1t=c  ⇒  t2−ct+1=0t + \frac1t = c \;\Rightarrow\; t^2 - ct + 1 = 0

Worked example

If 6(tan⁡θ+cot⁡θ)=136(\tan\theta + \cot\theta) = 13 with 0∘<θ<45∘0^\circ < \theta < 45^\circ, find sin⁡θ\sin\theta.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2021 · CDS (I) 2021 — Elementary Mathematics · Q56Moderate

Example 2 · Trigonometric Ratios and Identities · Trigonometric Equations

If sec⁡θ+cos⁡θ=52\sec\theta + \cos\theta = \frac{5}{2}, where 0≤θ≤90∘0 \le \theta \le 90^\circ, then what is the value of sin⁡2θ\sin^2\theta?

Both roots of tan θ + cot θ = c are real angles

tan⁡θ=43\tan\theta = \dfrac43 and tan⁡θ=34\tan\theta = \dfrac34 both solve 12(tan⁡θ+cot⁡θ)=2512(\tan\theta + \cot\theta) = 25; they are complementary angles. Only the stated range decides which one the question means — so a range like 45∘<θ<90∘45^\circ < \theta < 90^\circ is not decoration.
Drill 4 more on equations in a ratio and its reciprocal

Concept 3 of 4: Linear equations a sin θ + b cos θ = c

An equation like 8sin⁡θ−cos⁡θ=48\sin\theta - \cos\theta = 4 has two unknowns tied by sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1. Solve the linear equation for one of them, substitute into the identity, and a quadratic appears. One of its roots usually gives the wrong sign for the range and must go.

Definition

  • From asin⁡θ+bcos⁡θ=ca\sin\theta + b\cos\theta = c, express cos⁡θ\cos\theta in terms of sin⁡θ\sin\theta (or the reverse).
  • Substitute into sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 and solve the quadratic.
  • For each root, compute the other ratio and check its sign against the range. Squaring has introduced a spurious root.
  • If c2=a2+b2c^2 = a^2 + b^2, the equation is at its maximum and has one solution: sin⁡θ=ac\sin\theta = \dfrac{a}{c}, cos⁡θ=bc\cos\theta = \dfrac{b}{c}.

Substitute into the identity

asin⁡θ+bcos⁡θ=c,sin⁡2θ+cos⁡2θ=1a\sin\theta + b\cos\theta = c, \quad \sin^2\theta + \cos^2\theta = 1

Worked example

If 7sin⁡θ−cos⁡θ=57\sin\theta - \cos\theta = 5 with 0<θ<90∘0 < \theta < 90^\circ, find sin⁡θ\sin\theta.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2025 · CDS (II) 2025 — Elementary Mathematics · Q32Moderate

Example 3 · Trigonometric Ratios and Identities · Trigonometric Equations

If 8sin⁡θ−cos⁡θ=48\sin\theta - \cos\theta = 4, where 0<θ<π/20 < \theta < \pi/2, then what is cosec θ\mathrm{cosec}\,\theta equal to?

The quadratic always offers a root that fails

Substituting and squaring doubles the solutions. In the first quadrant both sine and cosine must be positive; the rejected root almost always gives a negative cosine. The paper prints its value among the options.
Drill 2 more on linear equations a sin θ + b cos θ = c

Concept 4 of 4: Systems in two or three angles

When the paper gives cos⁡(x+y)=0\cos(x + y) = 0 and sin⁡(x−y)=12\sin(x - y) = \dfrac12, each equation pins down one combination of the angles as a standard value. The trig is over after one line; what remains is solving two linear equations for xx and yy.

Definition

  • Turn each given value into an angle, using the range to choose it: cos⁡(x+y)=0\cos(x + y) = 0 with x+y∈[0,π]x + y \in [0, \pi] gives x+y=π2x + y = \dfrac{\pi}{2}.
  • Solve the resulting linear system.
  • For three angles given as B+C−AB + C - A, C+A−BC + A - B, A+B−CA + B - C, add the three combinations to get A+B+CA + B + C directly.

Sum of the three combinations

(B+C−A)+(C+A−B)+(A+B−C)=A+B+C(B + C - A) + (C + A - B) + (A + B - C) = A + B + C

Worked example

If sin⁡(A+B)=1\sin(A + B) = 1 and cos⁡(A−B)=32\cos(A - B) = \dfrac{\sqrt3}{2} with A>BA > B acute, find AA and BB.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2022 · CDS (I) 2022 — Elementary Mathematics · Q47Moderate

Example 4 · Trigonometric Ratios and Identities · Trigonometric Equations

If cos⁡(x+y)=0\cos(x + y) = 0 and sin⁡(x−y)=12\sin(x - y) = \frac{1}{2}, where x,y∈[0,π2]x, y \in \left[0, \frac{\pi}{2}\right], then what is the value of cot⁡(2x−y)\cot(2x - y)?

Use the range to choose the angle, not the calculator's first answer

sin⁡(A+B)=32\sin(A + B) = \dfrac{\sqrt3}{2} allows A+B=60∘A + B = 60^\circ or 120∘120^\circ. The condition that AA and BB are acute decides which, and the wrong one is always among the options.
Drill 1 more on systems in two or three 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 (4)

  • Reduce to one ratio and solve the quadratic

    The substitutions

    cos⁡2θ=1−sin⁡2θ,tan⁡2θ=sec⁡2θ−1\cos^2\theta = 1 - \sin^2\theta, \qquad \tan^2\theta = \sec^2\theta - 1
  • Equations in a ratio and its reciprocal

    Clearing the reciprocal

    t+1t=c  ⇒  t2−ct+1=0t + \frac1t = c \;\Rightarrow\; t^2 - ct + 1 = 0
  • Linear equations a sin θ + b cos θ = c

    Substitute into the identity

    asin⁡θ+bcos⁡θ=c,sin⁡2θ+cos⁡2θ=1a\sin\theta + b\cos\theta = c, \quad \sin^2\theta + \cos^2\theta = 1
  • Systems in two or three angles

    Sum of the three combinations

    (B+C−A)+(C+A−B)+(A+B−C)=A+B+C(B + C - A) + (C + A - B) + (A + B - C) = A + B + C

Watch out for (4)

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