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NotesMath AATopic 3.6
Unit 3 · Geometry & Trigonometry · Topic 3.6

IB Math AA — 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 exams and builds the foundation for connected topics across the syllabus.

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. 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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