M2 Trigonometry

2015-01-19 1:43 am
1. Solve √3 sec²θ + (√3 -1)tanθ - (1+√3) = 0, where 0 ≤ θ ≤ 2π

2. Let f(x) = [cos^(4)x - sin^(4)x] / (2cos²x + 5)
(a) Prove that f(x) = 1- 6/(2cos²+5).
(b) Hence, find the maximum and minimum values of f(x).

回答 (1)

2015-01-19 4:40 am
✔ 最佳答案
1. As sec²θ=1+tan²θ, so√3 sec²θ+(√3-1) tan θ-(1+√3)=0==> √3 (1+tan²θ)+(√3-1) tan θ-(1+√3)=0==> √3 tan²θ+(√3-1) tan θ-1=0==> (√3 tan θ-1)(tan θ+1)=0==> tan θ=1/√3 or -1==> θ=π/6, π+π/6, π-π/4, 2π-π/4==> θ=π/6, 4π/3, 7π/6, 7π/4
2(a). f(x)=(cos⁴x-sin⁴x) / (2cos²x+5)=(cos²x+sin²x)(cos²x-sin²x) / (2cos²x+5)=(cos²x-sin²x) / (2cos²x+5)=(cos²x-1+cos²x) / (2cos²x+5)=(2cos²x+5-6) / (2cos²x+5)=1-6/(2cos²x+5)
2(b). As 0 ≦ cos²x ≦ 1==> 2(0)+5 ≦ 2cos²x+5 ≦ 2(1)+5==> 6/5 ≧ 6/(2cos²x+5) ≧ 6/7==> 1-6/5 ≦ 1-6/(2cos²x+5) ≦ 1-6/7==> -1/5 ≦ f(x) ≦ 1/7∴ max value of f(x) is 1/7, min value of f(x) is -1/5.


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