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已知cos(π/4+θ)cos(π/4-θ)=1/4,则(sinθ)^4+(cosθ)^4
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已知cos(π/4+θ)cos(π/4-θ)=1/4,则(sinθ)^4+(cosθ)^4
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答案和解析
cos(π/4+θ)cos(π/4-θ)=1/4
左边按照积化和差公式：
1/2cos(π/4+θ-π/4+θ)+1/2cos(π/4+θ+π/4-θ) = 1/4
1/2cos(2θ)+1/2cos(π/2) = 1/4
1/2cos(2θ)+0 = 1/4
cos(2θ)=1/2
(sin2θ)^2=1-(cos2θ)^2=1-1/4=3/4
(sinθ)^4+(cosθ)^4= { (sinθ)^2+(cosθ)^2 }^2 - 2(sinθ)^2 (cosθ)^2 = 1 - 1/2(sin2θ)^2 = 1-1/2*3/4 = 5/8
左边按照积化和差公式：
1/2cos(π/4+θ-π/4+θ)+1/2cos(π/4+θ+π/4-θ) = 1/4
1/2cos(2θ)+1/2cos(π/2) = 1/4
1/2cos(2θ)+0 = 1/4
cos(2θ)=1/2
(sin2θ)^2=1-(cos2θ)^2=1-1/4=3/4
(sinθ)^4+(cosθ)^4= { (sinθ)^2+(cosθ)^2 }^2 - 2(sinθ)^2 (cosθ)^2 = 1 - 1/2(sin2θ)^2 = 1-1/2*3/4 = 5/8
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