36 exam-style questions on Pythagorean identity; double angle identities (sin, cos) — topic 3.6 of Geometry and Trigonometry in the IB Mathematics: Analysis and Approaches syllabus — with answers and explanations. Samples below; the full set is free with an account.
Which expression is equal to cos 2θ, written in terms of sin θ only?
Answer: 1 − 2 sin²θ
Substituting cos²θ = 1 − sin²θ into cos 2θ = cos²θ − sin²θ gives cos 2θ = 1 − 2 sin²θ; swapping the terms is a common sign error.
Given that cos θ = 1/3 and θ is acute, determine the exact value of sin 2θ.
Answer: 4√2/9
From sin²θ = 1 − 1/9 = 8/9 we get sin θ = 2√2/3 (positive since θ is acute), so sin 2θ = 2 sin θ cos θ = 2 × (2√2/3) × (1/3) = 4√2/9.
Simplify (1 − cos 2x) / sin 2x for values of x where the expression is defined.
Answer: tan x
Using 1 − cos 2x = 2 sin²x and sin 2x = 2 sin x cos x, the quotient becomes sin x / cos x = tan x.
Given that sin θ = 2/3 and π/2 < θ < π, determine the exact value of cos θ.
Answer: −√5/3
cos²θ = 1 − 4/9 = 5/9, and since θ is in the second quadrant cosine is negative, so cos θ = −√5/3.
Solve sin 2x = sin x for 0 < x < π/2.
Answer: x = π/3
sin 2x = 2 sin x cos x, and since sin x ≠ 0 on this interval we can divide by sin x to get cos x = 1/2, giving x = π/3.
Which expression gives sin²θ in terms of cos 2θ?
Answer: (1 − cos 2θ)/2
Rearranging cos 2θ = 1 − 2 sin²θ gives sin²θ = (1 − cos 2θ)/2; the version with a plus sign gives cos²θ instead.
That was 6 of 36 Pythagorean identity; double angle identities (sin, cos) questions.
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