These are comprehensive exercises on trigonometric equations and inequalities. Maximum-value problems are also included.
Example ① (Maximum Value by Substitution)
Find the maximum value of f(θ) = sinθcosθ + sinθ + cosθ（0 ≤ θ ≤ π）。
Answer:Let t = sinθ+cosθ（−1 ≤ t ≤ √2）。Then sinθcosθ = (t²−1)/2。
f = (t²−1)/2 + t = (t²+2t−1)/2 = ((t+1)²−2)/2。
The maximum occurs at t = √2 → f = (2+2√2−1)/2 = (1+2√2)/2
Find the maximum value of f(θ) = sinθcosθ + sinθ + cosθ（0 ≤ θ ≤ π）。
Answer:Let t = sinθ+cosθ（−1 ≤ t ≤ √2）。Then sinθcosθ = (t²−1)/2。
f = (t²−1)/2 + t = (t²+2t−1)/2 = ((t+1)²−2)/2。
The maximum occurs at t = √2 → f = (2+2√2−1)/2 = (1+2√2)/2
Example ②
Find the maximum and minimum values of f(θ) = cos²θ + sinθ（0 ≤ θ ≤ 2π）。
Answer:Let t = sinθ（−1 ≤ t ≤ 1）。Then f = 1−t²+t = −t²+t+1 = −(t−1/2)²+5/4。
Maximum value 5/4（t=1/2）, minimum value −1（t=−1）
Find the maximum and minimum values of f(θ) = cos²θ + sinθ（0 ≤ θ ≤ 2π）。
Answer:Let t = sinθ（−1 ≤ t ≤ 1）。Then f = 1−t²+t = −t²+t+1 = −(t−1/2)²+5/4。
Maximum value 5/4（t=1/2）, minimum value −1（t=−1）
Summary
- Basic forms: find the reference angle and list solutions by quadrant
- Quadratic forms: use identities to express everything in one function → factor
- Compound-angle forms: convert to r sin(θ+φ) → solve the basic equation
- Substitution forms: state the range of t before finding maximum and minimum values
Practice Problems
- Solve 2cos²θ − sinθ ≤ 1 for 0 ≤ θ < 2π.
- Find the maximum and minimum values of f(θ) = sinθ + cosθ for 0 ≤ θ ≤ π.
- Solve sin2θ > cosθ for 0 ≤ θ < 2π.
Answers and Explanations
- Answer:π/6 ≤ θ ≤ 5π/6 or θ = 3π/2
Explanation: 2(1−sin²θ)−sinθ≤1 → 2sin²θ+sinθ−1≥0 → (2sinθ−1)(sinθ+1)≥0。sinθ≥1/2 or sinθ≤−1。 - Answer:Maximum value √2（θ=π/4）, minimum value −1（θ=π）
Explanation: √2 sin(θ+π/4)。For θ∈[0,π], θ+π/4∈[π/4,5π/4]。The maximum is √2（θ=π/4）。At θ=π, sin π+cos π = −1。 - Answer:π/6 < θ < π/2 or 5π/6 < θ < π or π < θ < 3π/2
Explanation: 2sinθcosθ > cosθ → cosθ(2sinθ−1) > 0。Cases: cosθ>0 and sinθ>1/2, or cosθ<0 and sinθ<1/2。