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NDA Maths · Trigonometric Identities

Double, Triple & Half-Angle

Double-angle (the most-used), triple-angle, and half-angle formulas — plus the symmetric tricks like sin α + cos α = p that feed straight into sin 2α.

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Why this matters

Half of this subtopic's questions are HARD — the densest difficulty pocket in the chapter. Most resolve to picking the right form of cos 2A, knowing sin 3A = 3 sin A − 4 sin³A, or recognising a half-angle in 1 ± cos A.

Test yourself — a quick 12-question Trigonometric Identities recall quiz, instant score.

Concept 1 of 4: Double-angle formulas

Set B = A in the compound formulas. Cosine of a double angle has three interchangeable forms — the art is choosing the one that matches what you're given (sin only, cos only, or tan only).

Definition

  • sin⁡2A=2sin⁡Acos⁡A=2tan⁡A1+tan⁡2A\sin 2A=2\sin A\cos A=\dfrac{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=\dfrac{1-\tan^2 A}{1+\tan^2 A}.
  • tan⁡2A=2tan⁡A1−tan⁡2A\tan 2A=\dfrac{2\tan A}{1-\tan^2 A}. Also tan⁡A+cot⁡A=2sin⁡2A=2csc⁡2A\tan A+\cot A=\dfrac{2}{\sin 2A}=2\csc 2A.
  • The sibling, the most-used untaught line in the PYQ bank: cot⁡A−tan⁡A=cos⁡2A−sin⁡2Asin⁡Acos⁡A=2cot⁡2A\cot A-\tan A=\dfrac{\cos^2A-\sin^2A}{\sin A\cos A}=2\cot 2A. So tan⁡A+2tan⁡2A+4tan⁡4A+8cot⁡8A\tan A+2\tan 2A+4\tan 4A+8\cot 8A telescopes to cot⁡A\cot A (each 2kcot⁡2kA−2ktan⁡2kA=2k+1cot⁡2k+1A2^k\cot 2^kA-2^k\tan 2^kA=2^{k+1}\cot 2^{k+1}A), and tan⁡65°−tan⁡25°=2cot⁡50°=2tan⁡40°\tan 65°-\tan 25°=2\cot 50°=2\tan 40° settles tan⁡65°+2tan⁡45°−2tan⁡40°−tan⁡25°=2\tan 65°+2\tan 45°-2\tan 40°-\tan 25°=2.

Double-angle formulas

sin⁡2A=2sin⁡Acos⁡A,cos⁡2A=cos⁡2A−sin⁡2A=1−2sin⁡2A=2cos⁡2A−1,tan⁡2A=2tan⁡A1−tan⁡2A\sin 2A=2\sin A\cos A,\qquad \cos 2A=\cos^2 A-\sin^2 A=1-2\sin^2 A=2\cos^2 A-1,\qquad \tan 2A=\dfrac{2\tan A}{1-\tan^2 A}

Worked example

If tan⁡A=34\tan A=\tfrac34, find sin⁡2A\sin 2A.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2025 · Q45Moderate

Example 1 · Trigonometric Identities · Multiple and Half-Angle Formulas

Let 2sin⁡α+cos⁡α=22\sin\alpha + \cos\alpha = 2, where 0<α<90°0 < \alpha < 90°.
What is 2sin⁡2α+cos⁡2α2\sin 2\alpha + \cos 2\alpha equal to?

sin⁡2A=2sin⁡Acos⁡A\sin 2A=2\sin A\cos A — not 2sin⁡A2\sin A, and (sin⁡A)2≠sin⁡2A(\sin A)^2\neq\sin 2A

Two slips: dropping the cos⁡A\cos A (writing sin⁡2A=2sin⁡A\sin 2A=2\sin A), and confusing the double angle with a square (sin⁡2A\sin 2A is not sin⁡2A\sin^2 A). The identity is sin⁡2A=2sin⁡Acos⁡A\sin 2A=2\sin A\cos A. Check with A=30°A=30°: sin⁡60°=32\sin 60°=\tfrac{\sqrt3}{2}, while 2sin⁡30°=12\sin 30°=1 and sin⁡230°=14\sin^2 30°=\tfrac14 — all three differ.
Drill 11 more on double-angle formulas

Concept 2 of 4: Triple-angle formulas

The triple-angle identities turn sin 3A / cos 3A back into powers of sin A / cos A — and run in reverse to collapse "3 sin A − 4 sin³A" into a single sin 3A.

Definition

  • sin⁡3A=3sin⁡A−4sin⁡3A\sin 3A=3\sin A-4\sin^3 A.
  • cos⁡3A=4cos⁡3A−3cos⁡A\cos 3A=4\cos^3 A-3\cos A.
  • tan⁡3A=3tan⁡A−tan⁡3A1−3tan⁡2A\tan 3A=\dfrac{3\tan A-\tan^3 A}{1-3\tan^2 A}.

Triple-angle formulas

sin⁡3A=3sin⁡A−4sin⁡3A,cos⁡3A=4cos⁡3A−3cos⁡A,tan⁡3A=3tan⁡A−tan⁡3A1−3tan⁡2A\sin 3A=3\sin A-4\sin^3 A,\qquad \cos 3A=4\cos^3 A-3\cos A,\qquad \tan 3A=\dfrac{3\tan A-\tan^3 A}{1-3\tan^2 A}

Worked example

Simplify 3sin⁡20°−4sin⁡320°3\sin 20°-4\sin^3 20°.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2020 · Q25Moderate

Example 2 · Trigonometric Identities · Multiple and Half-Angle Formulas

What is sin⁡3x+cos⁡3x+4sin⁡3x−3sin⁡x+3cos⁡x−4cos⁡3x\sin 3x + \cos 3x + 4\sin^3 x - 3\sin x + 3\cos x - 4\cos^3 x equal to?

Watch the sign pattern: sin⁡3A=3sin⁡A−4sin⁡3A\sin 3A=3\sin A-4\sin^3 A but cos⁡3A=4cos⁡3A−3cos⁡A\cos 3A=4\cos^3 A-3\cos A

Students mix up the order and signs of the two triple-angle formulas. Sine starts with the linear term and subtracts the cube (3sin⁡A−4sin⁡3A3\sin A-4\sin^3 A); cosine starts with the cube and subtracts the linear term (4cos⁡3A−3cos⁡A4\cos^3 A-3\cos A). Writing sin⁡3A=4sin⁡3A−3sin⁡A\sin 3A=4\sin^3 A-3\sin A flips the whole sign — a guaranteed wrong answer.
Drill 5 more on triple-angle formulas

Concept 3 of 4: Half-angle formulas and 1 ± cos A / 1 ± sin A

Read the double-angle formulas backwards. The key recognitions: 1−cos⁡A=2sin⁡2A21-\cos A=2\sin^2\tfrac A2, 1+cos⁡A=2cos⁡2A21+\cos A=2\cos^2\tfrac A2, and 1±sin⁡A=(sin⁡A2±cos⁡A2)21\pm\sin A=(\sin\tfrac A2\pm\cos\tfrac A2)^2. These convert square roots into clean half-angle expressions.

Definition

  • 1−cos⁡A=2sin⁡2A21-\cos A=2\sin^2\tfrac A2,   1+cos⁡A=2cos⁡2A2\;1+\cos A=2\cos^2\tfrac A2.
  • tan⁡A2=sin⁡A1+cos⁡A=1−cos⁡Asin⁡A\tan\tfrac A2=\dfrac{\sin A}{1+\cos A}=\dfrac{1-\cos A}{\sin A}.
  • csc⁡A+cot⁡A=cot⁡A2\csc A+\cot A=\cot\tfrac A2,   csc⁡A−cot⁡A=tan⁡A2\;\csc A-\cot A=\tan\tfrac A2.
  • 1±sin⁡A=(sin⁡A2±cos⁡A2)21\pm\sin A=\left(\sin\tfrac A2\pm\cos\tfrac A2\right)^2 (mind the sign when taking the root).
  • Geometry that hides a half-angle: a chord subtending angle θ\theta at the centre of a circle of radius rr has length 2rsin⁡θ22r\sin\tfrac\theta2; the base of an isosceles triangle with legs aa and apex angle θ\theta is likewise 2asin⁡θ22a\sin\tfrac\theta2, and its base angles are 90°−θ290°-\tfrac\theta2.

Half-angle formulas

sin⁡A2=±1−cos⁡A2,cos⁡A2=±1+cos⁡A2,tan⁡A2=1−cos⁡Asin⁡A=sin⁡A1+cos⁡A\sin\tfrac A2=\pm\sqrt{\tfrac{1-\cos A}{2}},\qquad \cos\tfrac A2=\pm\sqrt{\tfrac{1+\cos A}{2}},\qquad \tan\tfrac A2=\dfrac{1-\cos A}{\sin A}=\dfrac{\sin A}{1+\cos A}

Worked example

Simplify 1−cos⁡2θsin⁡2θ\dfrac{1-\cos 2\theta}{\sin 2\theta}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2022 · Q48Moderate

Example 3 · Trigonometric Identities · Multiple and Half-Angle Formulas

What is the value of tan⁡ ⁣(3π8)\tan\!\left(\frac{3\pi}{8}\right)?

The ±\pm on a half-angle is fixed BY the quadrant of A/2A/2, not free

From sin⁡A2=±1−cos⁡A2\sin\tfrac A2=\pm\sqrt{\tfrac{1-\cos A}{2}} students grab the positive root automatically. But the sign is decided by which quadrant A/2A/2 lies in. E.g. if A=300°A=300° then A/2=150°A/2=150° (quadrant II), so sin⁡A2>0\sin\tfrac A2>0 but cos⁡A2<0\cos\tfrac A2<0. Likewise 1±sin⁡A=∣sin⁡A2±cos⁡A2∣\sqrt{1\pm\sin A}=\big|\sin\tfrac A2\pm\cos\tfrac A2\big| — take the modulus, then resolve the sign from the quadrant.
Drill 6 more on half-angle formulas and 1 ± cos a / 1 ± sin a

Concept 4 of 4: Symmetric tricks: sin ± cos, power reduction, sₙ patterns

A cluster of questions square a symmetric expression to expose a double angle (sin α + cos α squared gives 1 + sin 2α), reduce a fourth power to multiple angles, or chase patterns in tn=sin⁡nθ+cos⁡nθt_n=\sin^n\theta+\cos^n\theta.

Definition

  • Square the sum: (sin⁡α+cos⁡α)2=1+sin⁡2α(\sin\alpha+\cos\alpha)^2=1+\sin 2\alpha, (sin⁡α−cos⁡α)2=1−sin⁡2α(\sin\alpha-\cos\alpha)^2=1-\sin 2\alpha.
  • Power reduction: cos⁡4x=3+4cos⁡2x+cos⁡4x8\cos^4 x=\dfrac{3+4\cos 2x+\cos 4x}{8} (and similarly for sin⁡4\sin^4).
  • x+1x=2cos⁡θ⇒xn+1xn=2cos⁡nθx+\tfrac1x=2\cos\theta\Rightarrow x^n+\tfrac{1}{x^n}=2\cos n\theta (De Moivre flavour).

Worked example

If sin⁡α+cos⁡α=p\sin\alpha+\cos\alpha=p, express sin⁡2α\sin 2\alpha in terms of pp.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2019 · Q38Moderate

Example 4 · Trigonometric Identities · Multiple and Half-Angle Formulas

If sin⁡α+cos⁡α=p\sin\alpha + \cos\alpha = p, then what is cos⁡2(2α)\cos^2(2\alpha) equal to?
Drill 5 more on symmetric tricks: sin ± cos, power reduction, sₙ patterns

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)

  • Double-angle formulas

    Double-angle formulas

    sin⁡2A=2sin⁡Acos⁡A,cos⁡2A=cos⁡2A−sin⁡2A=1−2sin⁡2A=2cos⁡2A−1,tan⁡2A=2tan⁡A1−tan⁡2A\sin 2A=2\sin A\cos A,\qquad \cos 2A=\cos^2 A-\sin^2 A=1-2\sin^2 A=2\cos^2 A-1,\qquad \tan 2A=\dfrac{2\tan A}{1-\tan^2 A}
  • Triple-angle formulas

    Triple-angle formulas

    sin⁡3A=3sin⁡A−4sin⁡3A,cos⁡3A=4cos⁡3A−3cos⁡A,tan⁡3A=3tan⁡A−tan⁡3A1−3tan⁡2A\sin 3A=3\sin A-4\sin^3 A,\qquad \cos 3A=4\cos^3 A-3\cos A,\qquad \tan 3A=\dfrac{3\tan A-\tan^3 A}{1-3\tan^2 A}
  • Half-angle formulas and 1 ± cos A / 1 ± sin A

    Half-angle formulas

    sin⁡A2=±1−cos⁡A2,cos⁡A2=±1+cos⁡A2,tan⁡A2=1−cos⁡Asin⁡A=sin⁡A1+cos⁡A\sin\tfrac A2=\pm\sqrt{\tfrac{1-\cos A}{2}},\qquad \cos\tfrac A2=\pm\sqrt{\tfrac{1+\cos A}{2}},\qquad \tan\tfrac A2=\dfrac{1-\cos A}{\sin A}=\dfrac{\sin A}{1+\cos A}

Watch out for (3)

  • sin⁡2A=2sin⁡Acos⁡A\sin 2A=2\sin A\cos A — not 2sin⁡A2\sin A, and (sin⁡A)2≠sin⁡2A(\sin A)^2\neq\sin 2A
    → Double-angle formulas
  • Watch the sign pattern: sin⁡3A=3sin⁡A−4sin⁡3A\sin 3A=3\sin A-4\sin^3 A but cos⁡3A=4cos⁡3A−3cos⁡A\cos 3A=4\cos^3 A-3\cos A
    → Triple-angle formulas
  • The ±\pm on a half-angle is fixed BY the quadrant of A/2A/2, not free
    → Half-angle formulas and 1 ± cos A / 1 ± sin A

Test yourself on Trigonometric Identities

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