Setting α = β in the addition formulas gives the double-angle formulas.
- sin 2α = 2 sin α cos α
- cos 2α = cos²α − sin²α = 2cos²α − 1 = 1 − 2sin²α
- tan 2α = 2 tan α/(1 − tan²α)
Example ①
When sin θ = 3/5 (an acute angle), find sin 2θ and cos 2θ.
Answer:cos θ = 4/5。
sin 2θ = 2(3/5)(4/5) = 24/25
cos 2θ = (4/5)² − (3/5)² = 16/25 − 9/25 = 7/25
When sin θ = 3/5 (an acute angle), find sin 2θ and cos 2θ.
Answer:cos θ = 4/5。
sin 2θ = 2(3/5)(4/5) = 24/25
cos 2θ = (4/5)² − (3/5)² = 16/25 − 9/25 = 7/25
Example ②
Express sin 3θ as a polynomial in sin θ.
Answer:sin 3θ = sin(2θ+θ) = sin2θcosθ + cos2θsinθ
= 2sinθcos²θ + (1−2sin²θ)sinθ = 3sinθ − 4sin³θ
Express sin 3θ as a polynomial in sin θ.
Answer:sin 3θ = sin(2θ+θ) = sin2θcosθ + cos2θsinθ
= 2sinθcos²θ + (1−2sin²θ)sinθ = 3sinθ − 4sin³θ
Key Point
- Use the three forms of cos 2α appropriately (when expressing it using only one squared function)
- sin 2α = 2sinαcosα frequently appears in integration as well
Practice Problems
- When cos θ = −1/3（π/2 ≤ θ ≤ π）, find sin 2θ and cos 2θ.
- When tan θ = 3, find tan 2θ.
- Express cos 3θ as a polynomial in cos θ.
Answers and Explanations
- Answer:sin θ = 2√2/3、sin 2θ = −4√2/9、cos 2θ = −7/9
Explanation: In the second quadrant, sin > 0. sin 2θ = 2(2√2/3)(−1/3). cos 2θ = 1/9−8/9. - Answer:−3/4
Explanation: tan 2θ = 2×3/(1−9) = 6/(−8) = −3/4. - Answer:4cos³θ − 3cosθ
Explanation: cos3θ = cos(2θ+θ) = cos2θcosθ−sin2θsinθ = (2cos²θ−1)cosθ−2sinθcosθ·sinθ = 4cos³θ−3cosθ.