PYQ Vault
Ebooks

SSC CGL Maths · Trigonometric Ratios and Identities

Compound and Multiple Angles

The formulas for sin, cos and tan of A ± B, the double, half and triple angle forms that follow from them, and the rules that turn products into sums.

All Trigonometric Ratios and Identities notes

Why this matters

28 questions in the bank, and 16 of them are MODERATE, the highest share of any page. They test two things: recalling the right formula (several simply ask which option equals sin(A + B) or tan of a half angle) and recognising a printed expression as one side of a formula, so that it collapses to a single value such as sin 30° or cos 90°.

Concept 1 of 3: The addition and subtraction formulas

sin(A + B) is not sin A + sin B. Each formula mixes the ratios of both angles, and once you have them you can reach angles off the table, such as 15° = 45° − 30° and 75° = 45° + 30°.

Definition

  • sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B: the sign inside matches the sign outside.
  • cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B: the sign flips.
  • tan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}.
  • Values worth remembering: sin⁡15∘=6−24\sin 15^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}, cos⁡15∘=6+24\cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}, tan⁡15∘=2−3\tan 15^\circ = 2 - \sqrt{3}, tan⁡75∘=2+3\tan 75^\circ = 2 + \sqrt{3}. sin 75° and cos 15° are equal (co-functions).
  • Fast route: read the printed expression backwards. sin⁡20∘cos⁡25∘+cos⁡20∘sin⁡25∘\sin 20^\circ\cos 25^\circ + \cos 20^\circ\sin 25^\circ is sin⁡45∘\sin 45^\circ at sight.
  • If A+B=45∘A + B = 45^\circ, then tan⁡A+tan⁡B+tan⁡Atan⁡B=1\tan A + \tan B + \tan A\tan B = 1; if A−B=45∘A - B = 45^\circ, then tan⁡A−tan⁡B−tan⁡Atan⁡B=1\tan A - \tan B - \tan A\tan B = 1. Both come from tan⁡45∘=1\tan 45^\circ = 1.

Compound angles

sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡Bcos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡Btan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B \qquad \cos(A \pm B) = \cos A\cos B \mp \sin A\sin B \qquad \tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}

Worked example

Find the exact value of cos⁡105∘\cos 105^\circ.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 18 Sep 2025, Shift 2 · Q64Easy

Example 1 · Trigonometric Ratios and Identities · Compound and Multiple Angles

If sin⁡A=35\sin A = \frac{3}{5} and cos⁡B=1213\cos B = \frac{12}{13}, AA and B∈(0,π2)B \in \left(0, \frac{\pi}{2}\right), find sin⁡(A+B)\sin(A + B).

Keeping the sign in cos(A + B)

cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A + B) = \cos A\cos B - \sin A\sin B, with a minus. The plus version is cos⁡(A−B)\cos(A - B), and SSC prints both as options.

Splitting tan over a sum

tan⁡(A+B)\tan(A + B) is not tan⁡A+tan⁡B\tan A + \tan B. With A = B = 30°: tan⁡60∘=3≈1.73\tan 60^\circ = \sqrt{3} \approx 1.73, while 2tan⁡30∘≈1.152\tan 30^\circ \approx 1.15.
Drill 11 more on the addition and subtraction formulas

Concept 2 of 3: Double, half and triple angle formulas

Put B = A in the addition formulas and the double-angle forms drop out. Read them the other way and they halve an angle, which is how 1−cos⁡θ1 - \cos\theta and 1+cos⁡θ1 + \cos\theta turn into squares.

Definition

  • sin⁡2A=2sin⁡Acos⁡A=2tan⁡A1+tan⁡2A\sin 2A = 2\sin A\cos A = \frac{2\tan A}{1 + \tan^2 A}.
  • cos⁡2A=cos⁡2A−sin⁡2A=1−2sin⁡2A=2cos⁡2A−1=1−tan⁡2A1+tan⁡2A\cos 2A = \cos^2 A - \sin^2 A = 1 - 2\sin^2 A = 2\cos^2 A - 1 = \frac{1 - \tan^2 A}{1 + \tan^2 A}.
  • So 1−cos⁡2A=2sin⁡2A1 - \cos 2A = 2\sin^2 A, 1+cos⁡2A=2cos⁡2A1 + \cos 2A = 2\cos^2 A, and 1−cos⁡2A1+cos⁡2A=tan⁡2A\frac{1 - \cos 2A}{1 + \cos 2A} = \tan^2 A.
  • Half angle: tan⁡A2=sin⁡A1+cos⁡A=1−cos⁡Asin⁡A=1−cos⁡A1+cos⁡A\tan\frac{A}{2} = \frac{\sin A}{1 + \cos A} = \frac{1 - \cos A}{\sin A} = \sqrt{\frac{1 - \cos A}{1 + \cos A}} for an acute A.
  • Triple angle: sin⁡3A=3sin⁡A−4sin⁡3A\sin 3A = 3\sin A - 4\sin^3 A and cos⁡3A=4cos⁡3A−3cos⁡A\cos 3A = 4\cos^3 A - 3\cos A.
  • Nested roots: 2+2cos⁡2A=2cos⁡A\sqrt{2 + 2\cos 2A} = 2\cos A when cos⁡A≥0\cos A \ge 0, so each square root halves the angle.
  • (sin⁡A+cos⁡A)2=1+sin⁡2A(\sin A + \cos A)^2 = 1 + \sin 2A, so a given sum of sin and cos gives sin 2A directly.
  • Fast route: to choose between formula options, test A = 60°, where the double angle 120° and the half angle 30° are both known. Avoid A = 90°, where several half-angle forms all equal 1.

Multiple angles

sin⁡2A=2sin⁡Acos⁡A=2tan⁡A1+tan⁡2Acos⁡2A=1−2sin⁡2A=2cos⁡2A−1=1−tan⁡2A1+tan⁡2Atan⁡A2=sin⁡A1+cos⁡Asin⁡3A=3sin⁡A−4sin⁡3Acos⁡3A=4cos⁡3A−3cos⁡A\sin 2A = 2\sin A\cos A = \frac{2\tan A}{1 + \tan^2 A} \qquad \cos 2A = 1 - 2\sin^2 A = 2\cos^2 A - 1 = \frac{1 - \tan^2 A}{1 + \tan^2 A} \qquad \tan\frac{A}{2} = \frac{\sin A}{1 + \cos A} \qquad \sin 3A = 3\sin A - 4\sin^3 A \qquad \cos 3A = 4\cos^3 A - 3\cos A

Worked example

If tan⁡A=2\tan A = 2, find sin⁡2A\sin 2A and cos⁡2A\cos 2A.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 20 Sep 2025, Shift 1 · Q74Moderate

Example 2 · Trigonometric Ratios and Identities · Compound and Multiple Angles

If sin⁡x=0.6\sin x = 0.6 and x∈(0,π2)x \in \left(0, \frac{\pi}{2}\right), find sin⁡2x+cos⁡2x\sin 2x + \cos 2x.

Doubling the ratio instead of the angle

sin⁡2A\sin 2A is not 2sin⁡A2\sin A. At A = 60°, sin⁡120∘=32\sin 120^\circ = \frac{\sqrt{3}}{2} while 2sin⁡60∘=32\sin 60^\circ = \sqrt{3}.

Turning the half-angle fraction upside down

sin⁡A1+cos⁡A\frac{\sin A}{1 + \cos A} is tan⁡A2\tan\frac{A}{2}, but sin⁡A1−cos⁡A\frac{\sin A}{1 - \cos A} is cot⁡A2\cot\frac{A}{2}. At A = 60° the first is 13\frac{1}{\sqrt{3}} and the second 3\sqrt{3}.
Drill 12 more on double, half and triple angle formulas

Concept 3 of 3: Products and sums of sines and cosines

Add or subtract the two addition formulas and the cross terms cancel, leaving a product equal to a sum. Read backwards, a sum of two sines or cosines becomes a product, which is what makes long sums factor.

Definition

  • 2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\sin B = \cos(A - B) - \cos(A + B).
  • 2cos⁡Acos⁡B=cos⁡(A−B)+cos⁡(A+B)2\cos A\cos B = \cos(A - B) + \cos(A + B).
  • 2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2\sin A\cos B = \sin(A + B) + \sin(A - B).
  • Backwards: sin⁡C+sin⁡D=2sin⁡C+D2cos⁡C−D2\sin C + \sin D = 2\sin\frac{C + D}{2}\cos\frac{C - D}{2} and cos⁡C+cos⁡D=2cos⁡C+D2cos⁡C−D2\cos C + \cos D = 2\cos\frac{C + D}{2}\cos\frac{C - D}{2}.
  • A product of a sum and a difference: cos⁡(A+B)cos⁡(A−B)=cos⁡2A−sin⁡2B\cos(A + B)\cos(A - B) = \cos^2 A - \sin^2 B.
  • Fast route: check a formula option at A = B. Then sin⁡Asin⁡B=sin⁡2A\sin A\sin B = \sin^2 A, and only the right option gives 12(1−cos⁡2A)=sin⁡2A\frac{1}{2}(1 - \cos 2A) = \sin^2 A.

Product and sum formulas

2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2cos⁡Acos⁡B=cos⁡(A−B)+cos⁡(A+B)2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)sin⁡C+sin⁡D=2sin⁡C+D2cos⁡C−D2cos⁡C+cos⁡D=2cos⁡C+D2cos⁡C−D22\sin A\sin B = \cos(A - B) - \cos(A + B) \qquad 2\cos A\cos B = \cos(A - B) + \cos(A + B) \qquad 2\sin A\cos B = \sin(A + B) + \sin(A - B) \qquad \sin C + \sin D = 2\sin\tfrac{C + D}{2}\cos\tfrac{C - D}{2} \qquad \cos C + \cos D = 2\cos\tfrac{C + D}{2}\cos\tfrac{C - D}{2}

Worked example

Find the value of 2sin⁡75∘sin⁡15∘2\sin 75^\circ\sin 15^\circ.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 18 Sep 2024, Shift 3 · Q55Easy

Example 3 · Trigonometric Ratios and Identities · Compound and Multiple Angles

SinASinB=\text{SinASinB} = ___________.

Reversing the order in 2 sin A sin B

2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\sin B = \cos(A - B) - \cos(A + B). Written the other way round, cos⁡(A+B)−cos⁡(A−B)\cos(A + B) - \cos(A - B), it is the negative of the right answer, and that form is always offered.
Drill 2 more on products and sums of sines and cosines

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 (3)

  • The addition and subtraction formulas

    Compound angles

    sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡Bcos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡Btan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B \qquad \cos(A \pm B) = \cos A\cos B \mp \sin A\sin B \qquad \tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}
  • Double, half and triple angle formulas

    Multiple angles

    sin⁡2A=2sin⁡Acos⁡A=2tan⁡A1+tan⁡2Acos⁡2A=1−2sin⁡2A=2cos⁡2A−1=1−tan⁡2A1+tan⁡2Atan⁡A2=sin⁡A1+cos⁡Asin⁡3A=3sin⁡A−4sin⁡3Acos⁡3A=4cos⁡3A−3cos⁡A\sin 2A = 2\sin A\cos A = \frac{2\tan A}{1 + \tan^2 A} \qquad \cos 2A = 1 - 2\sin^2 A = 2\cos^2 A - 1 = \frac{1 - \tan^2 A}{1 + \tan^2 A} \qquad \tan\frac{A}{2} = \frac{\sin A}{1 + \cos A} \qquad \sin 3A = 3\sin A - 4\sin^3 A \qquad \cos 3A = 4\cos^3 A - 3\cos A
  • Products and sums of sines and cosines

    Product and sum formulas

    2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2cos⁡Acos⁡B=cos⁡(A−B)+cos⁡(A+B)2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)sin⁡C+sin⁡D=2sin⁡C+D2cos⁡C−D2cos⁡C+cos⁡D=2cos⁡C+D2cos⁡C−D22\sin A\sin B = \cos(A - B) - \cos(A + B) \qquad 2\cos A\cos B = \cos(A - B) + \cos(A + B) \qquad 2\sin A\cos B = \sin(A + B) + \sin(A - B) \qquad \sin C + \sin D = 2\sin\tfrac{C + D}{2}\cos\tfrac{C - D}{2} \qquad \cos C + \cos D = 2\cos\tfrac{C + D}{2}\cos\tfrac{C - D}{2}

Watch out for (5)

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.

Take the Trigonometric Ratios and Identities chapter testDrill all 28 questions