Double the angle, not the value: sin 2θ is NOT 2 sin θ. The correct double-angle formula is sin 2θ = 2 sin θ cos θ.
IB-style question — find sin 2θ
Given sin θ = 3/5 and cos θ = 4/5, find sin 2θ.
Step by step
- Apply the formula.
Final answer
sin 2θ = 24/25.
A very common slip: Writing sin 2θ = 2 sin θ loses the cos θ factor — always include both.
One formula, three faces: cos 2θ = cos²θ − sin²θ = 1 − 2sin²θ = 2cos²θ − 1. The three are equal (via sin²+cos²=1); pick whichever matches what you know.
Which form?: Know only sin? Use 1 − 2sin²θ. Know only cos? Use 2cos²θ − 1. Know both? Any form works.
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Get sin θ and cos θ first: To find sin 2θ or cos 2θ, first find sin θ and cos θ (often via the Pythagorean identity), then substitute into the double-angle formula.
IB-style question — chain the identities
Given cos θ = 4/5 with θ acute, find cos 2θ.
Step by step
- Use the cos-only form.
Final answer
cos 2θ = 7/25.
Pick the form that avoids extra work: Here using 2cos²θ − 1 means you never need sin θ at all — choose the form that uses what you're given.
Spot the double-angle pattern: When an expression has 2 sin θ cos θ or cos²θ − sin²θ, replace it with sin 2θ or cos 2θ. Difference-of-squares + the Pythagorean identity often reveal it.
IB-style question — the audited identity
Show that cos⁴θ − sin⁴θ = cos 2θ.
Step by step
- Difference of two squares.
- The second bracket is 1; the first is cos 2θ.
Final answer
So cos⁴θ − sin⁴θ ≡ cos 2θ. (the audited 'show that')
Factor, then identify: Factor where you can; an expression like cos²θ − sin²θ is exactly cos 2θ.