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
- any angle at all — acute, obtuse, negative
- shorthand for (\sin\theta)^{2}
✌️ Double-angle formulas
- double the angle — NOT double the value
- 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:
Rearrange the Pythagorean identity.
Root it; θ is acute, so take the positive root.
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:
Find cos θ from the identity.
Quadrant 2 ⇒ cos θ is negative.
Apply sin 2θ = 2 sin θ cos θ.
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.
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.