若cos(π/4+θ)cos(π/4-θ)=1/4,则sin4θ+cos4θ等于
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若cos(π/4+θ)cos(π/4-θ)=1/4,则sin4θ+cos4θ等于若cos(π/4+θ)cos(π/4-θ)=1/4,则sin4θ+cos4θ等于若cos(π/4+θ)cos(π/4-θ
若cos(π/4+θ)cos(π/4-θ)=1/4,则sin4θ+cos4θ等于
若cos(π/4+θ)cos(π/4-θ)=1/4,则sin4θ+cos4θ等于
若cos(π/4+θ)cos(π/4-θ)=1/4,则sin4θ+cos4θ等于
根据：2cosαcosβ = cos(α + β) + cos(α - β)
左端 = [cos(π/2) + cos(2θ)] / 2 = cos(2θ) / 2 = 1/4 = 右端
cos2θ = 1/2 ； sin2θ = √(1 - cos²2θ) = ±√3/2 ------ 或者 2θ = ±π/3
sin4θ = 2sin2θcos2θ = ±√3/2 ①
cos4θ = (2cos²2θ - 1) = - 1/2 ②
sin4θ + cos4θ = ① + ② = - 1/2 ±√3/2 = - (1 ±√3) / 2 ---------- 这就是答案
或者sin4θ + cos4θ = sin(±2π/3) + cos(±2π/3) = ±√3/2 - 1/2
sqrt(3)/2 - 1/2或
-sqrt(3)/2 - 1/2;
cos(π/4+θ)cos(π/4-θ)=sin(π/4-θ)cos(π/4-θ)=1/2sin(π/2-2θ)=1/4
sin(π/2-2θ)=1/2
cos2θ=1/2 sin2θ=±√3/2
sin4θ=2*sin2θ*cos2θ=±√3/2
cos4θ=cos^2(2θ)-sin^2(2θ)=-1/2
sin4θ+cos4θ等于-1/2±√3/2
对不起我想提问
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