These exercises comprehensively review the addition theorems, double-angle formulas, half-angle formulas, and compound-angle transformations.
Example ①
Find the value of sin 15° using the half-angle formula.
Answer:sin²15° = (1−cos30°)/2 = (1−√3/2)/2 = (2−√3)/4。
Since sin 15° > 0, sin 15° = √(2−√3)/2
Find the value of sin 15° using the half-angle formula.
Answer:sin²15° = (1−cos30°)/2 = (1−√3/2)/2 = (2−√3)/4。
Since sin 15° > 0, sin 15° = √(2−√3)/2
Example ②
Find the maximum value of 3sinθ − 4cosθ.
Answer:r = √(9+16) = √25 = 5
Find the maximum value of 3sinθ − 4cosθ.
Answer:r = √(9+16) = √25 = 5
Summary
- Six addition theorems (sin±, cos±, tan±)
- Double-angle formulas: sin2α=2sinαcosα and the three forms of cos2α
- Half-angle formulas: sin²(α/2)=(1−cosα)/2、cos²(α/2)=(1+cosα)/2
- Compound angles: r=√(a²+b²), and φ is determined using both cos and sin
Practice Problems
- When cos θ = 2/3, find cos 2θ and cos 4θ.
- Let sinθ + cosθ = t. Using sin²θ + cos²θ = 1, find the range of possible values of t.
- Find the maximum value of 2sinθ − cosθ and the value of θ when it occurs (for general θ).
Answers and Explanations
- Answer:cos2θ = 2(4/9)−1 = −1/9。cos4θ = 2(1/81)−1 = −79/81
Explanation: cos2θ = 2cos²θ−1。cos4θ = 2cos²2θ−1。 - Answer:−√2 ≤ t ≤ √2
Explanation: Since t = √2 sin(θ+π/4), |t| ≤ √2。 - Answer:Maximum value √5 (when θ = π/2 + arctan(1/2) + 2nπ)
Explanation: r = √(4+1) = √5。