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c059741
NotesMath AA HLTopic 3.6
Unit 3 · Geometry & Trigonometry · Topic 3.6

IB Math AA HL — Identities & double angles

Topic 3.6 of IB Mathematics: Analysis and Approaches covers Identities & double angles, which is part of Unit 3: Geometry & Trigonometry. Students explore key concepts including Pythagorean identity, Double angles. A strong understanding of identities & double angles is essential for IB Math AA HL exams and builds the foundation for connected topics across the syllabus.

Higher Level students should use this topic hub as a map: start with the shared sub-topics, then follow the HL-only extensions and exam-skill links where this topic asks for deeper analysis.

Exam technique guidePractice questions

Key concepts in Identities & double angles

Key Idea: These are the trig identities that let you swap one ratio for another and rewrite double angles — the engine behind simplify, show that, and find the exact value questions on Paper 1 (non-calculator).

🔺 The Pythagorean identity

sin⁡2θ+cos⁡2θ=1\sin^{2}\theta + \cos^{2}\theta = 1sin2θ+cos2θ=1
θ\thetaθ
any angle at all — acute, obtuse, negative
sin⁡2θ\sin^{2}\thetasin2θ
shorthand for (\sin\theta)^{2}
You have…Rearranged formUsed to…
sin θcos²θ = 1 − sin²θfind cos θ (then ±√, sign by quadrant)
cos θsin²θ = 1 − cos²θfind sin θ (then ±√, sign by quadrant)
1 − sin²θ or 1 − cos²θswap for cos²θ / sin²θcancel terms in a show that

✌️ Double-angle formulas

sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin\theta\cos\thetasin2θ=2sinθcosθ
2θ2\theta2θ
double the angle — NOT double the value
cos⁡2θ=cos⁡2θ−sin⁡2θ=1−2sin⁡2θ=2cos⁡2θ−1\cos 2\theta = \cos^{2}\theta - \sin^{2}\theta = 1 - 2\sin^{2}\theta = 2\cos^{2}\theta - 1cos2θ=cos2θ−sin2θ=1−2sin2θ=2cos2θ−1
===
all three forms are equal via sin²+cos²=1
Know only sin θ? Use 1 − 2sin²θ. Know only cos θ? Use 2cos²θ − 1. Know both? Any form works. Choosing well means you never compute the ratio you weren't given.

✏️ IB-style worked examples

IB-style question — find sin θ from cos θ

Given cos θ = 3/4 with θ acute, find the exact value of sin θ.

Step by step:

  1. Rearrange the Pythagorean identity.

    sin⁡2θ=1−(34)2=716\sin^{2}\theta = 1 - \left(\tfrac{3}{4}\right)^{2} = \tfrac{7}{16}sin2θ=1−(43​)2=167​
  2. Root it; θ is acute, so take the positive root.

    sin⁡θ=74\sin\theta = \tfrac{\sqrt{7}}{4}sinθ=47​​
Final answer:

sin θ = √7 / 4

IB-style question — find sin 2θ given a quadrant

Given sin θ = 5/13 with θ in the second quadrant, find the exact value of sin 2θ.

Step by step:

  1. Find cos θ from the identity.

    cos⁡2θ=1−25169=144169\cos^{2}\theta = 1 - \tfrac{25}{169} = \tfrac{144}{169}cos2θ=1−16925​=169144​
  2. Quadrant 2 ⇒ cos θ is negative.

    cos⁡θ=−1213\cos\theta = -\tfrac{12}{13}cosθ=−1312​
  3. Apply sin 2θ = 2 sin θ cos θ.

    sin⁡2θ=2(513)(−1213)=−120169\sin 2\theta = 2\left(\tfrac{5}{13}\right)\left(-\tfrac{12}{13}\right) = -\tfrac{120}{169}sin2θ=2(135​)(−1312​)=−169120​
Final answer:

sin 2θ = −120/169

Important: sin 2θ ≠ 2 sin θ and sin²θ ≠ sin θ². Always keep the cos θ factor in 2 sin θ cos θ, and read sin²θ as (sin θ)². Dropping either loses the whole mark.

Tap each card to reveal the answer.

Simplify 1 − sin²θ cos²θ — the rearranged Pythagorean identity.

cos θ = 8/17, θ acute — find sin θ 15/17 — sin²θ = 1 − 64/289 = 225/289, positive root.

sin θ = 1/2, cos θ = √3/2 — find sin 2θ √3/2 — 2 × ½ × √3/2 = √3/2.

cos θ = 3/5, θ acute — find cos 2θ −7/25 — use 2cos²θ − 1 = 18/25 − 1.

Show that 1 − cos 2θ = 2 sin²θ Replace cos 2θ with 1 − 2sin²θ: 1 − (1 − 2sin²θ) = 2 sin²θ.

Which cos 2θ form if you only know sin θ? 1 − 2sin²θ — it needs sin θ only.

Exam Tips

  • sin²θ + cos²θ = 1 holds for every angle — rearrange to swap one squared ratio for the other.
  • Find a ratio with ±√(1 − …), then fix the sign from the quadrant (acute ⇒ positive).
  • sin 2θ = 2 sin θ cos θ — never drop the cos θ; sin 2θ is not 2 sin θ.
  • For cos 2θ pick the form matching your info: 1 − 2sin²θ (sin only) or 2cos²θ − 1 (cos only).
  • For exact values, find sin θ and cos θ first, then substitute — all by hand on Paper 1.

What you'll learn in Topic 3.6

  • 3.6.1 Pythagorean identity
  • 3.6.2 Double angles
Suggested study order: Read the notes for each sub-topic below → test yourself with flashcards → attempt practice questions → review exam technique.

Study resources — 3.6 Identities & double angles

3.6.1

Pythagorean identity

Notes
3.6.2

Double angles

Notes

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Topic 3.6 Identities & double angles forms a core part of Unit 3: Geometry & Trigonometry in IB Math AA HL. Mastering these concepts will strengthen your understanding of connected topics across the syllabus and prepare you for exam questions that require analysis, evaluation, and real-world application.

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