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Principle deep dive

Double / half-angle formulas

sin 2A = 2 sin A cos A and cousins. 44% HARD overall — second-hardest principle in the bank after Sine/Cosine rules. Splits across Trig Identities Multi/Half-Angle and Properties of Triangle.

questions in the bank
41
tagged HARD
44%
chapter spread
2
worked examples below
4

When to reach for it

The question contains sin 2θ, cos 2θ, tan 2θ — or an expression that reduces to one of them after simplification.

Why this principle matters

Three formulas, three sign conventions, the whole chapter: sin 2A = 2 sin A cos A; cos 2A = cos²A − sin²A = 1 − 2sin²A = 2cos²A − 1; tan 2A = 2 tan A / (1 − tan²A). The three forms of cos 2A matter: each is the right tool for a different shape of input.

Double angle is the bank's hardest principle by HARD-rate (50%). The difficulty isn't the formula — it's recognising when an expression is secretly a double angle in disguise. (1 − tan²θ) / (1 + tan²θ) is cos 2θ. 2 sin θ cos θ is sin 2θ. cos²θ − sin²θ is cos 2θ. Recognising 'this is double angle' is half the work.

Half-angle is the same principle in reverse. cos A = 1 − 2sin²(A/2) gives sin(A/2) = √[(1 − cos A) / 2]. Useful for problems where you have cos A and need sin(A/2) or vice versa. NDA's chord-length-from-central-angle question is a half-angle in disguise.

4 worked examples from the bank

Each example demonstrates the principle on a real past-year question. Click to reveal the answer, then the solution.

Example 1Trigonometric IdentitiesEASY
If tan⁡A=17\tan A = \frac{1}{7}, then what is cos⁡2A\cos 2A equal to?

[Q32 · Apr · 2021]

Example 2Trigonometric IdentitiesEASY
If x+1x=2cos⁡θx+\dfrac{1}{x}=2\cos\theta, then what is x3+1x3x^3+\dfrac{1}{x^3} equal to?

[Q48 · Sep · 2025]

Example 3Trigonometric IdentitiesMODERATE
If α+β=π4\alpha+\beta=\frac{\pi}{4} and 2tan⁡α=12\tan\alpha=1, then what is tan⁡2β\tan2\beta equal to?

[Q28 · Apr · 2022]

Example 4Trigonometric IdentitiesHARD
What is the value of cos⁡47π8+cos⁡45π8\cos^4\frac{7\pi}{8}+\cos^4\frac{5\pi}{8}?

[Q32 · Apr · 2022]

Variants to recognise

Same principle, different surfaces. Pattern-match these on test day.

  • sin 2A = 2 sin A cos A

    Use the reverse direction when you see 2 sin A cos B — it collapses to sin(A + B) − sin(A − B) products.

  • cos 2A — three forms

    cos²A − sin²A (works with both); 1 − 2 sin²A (use when you know sin A); 2 cos²A − 1 (use when you know cos A).

  • tan 2A = 2 tan A / (1 − tan²A)

    Solve for tan A given tan 2A → quadratic in tan A. Sign of tan A by quadrant determines the root.

  • Half-angle from cos A

    sin(A/2) = √[(1 − cos A) / 2], cos(A/2) = √[(1 + cos A) / 2]. Sign depends on which quadrant A/2 lives in.

Drill every double / half-angle formulas question

41 questions from the bank — paginated, with cart and Word-export support.

Drill the 41 questions

Related principles

Often combined with this one — drill these next if you found the examples above tractable.