若sin^2θ+sinθ=1，则cos^2θ+cos^6θ=的值为多少？

数学若(1+sinθ+cosθ)

(1+sinθ+cosθ)/(1+sinθ-cosθ)=1/2 解：[(1+cost)+sint]/[(1-cost)+sint]=1/2 左边=[2(cos(t/2))^2+2sin(t/2)cos(t/2)] /[2(sin(t/2))^2+2sin(t/2)cos(t/2)] =2cos(t/2)[cos(t/2)+sin(t/2)] /{2sin(t/2)[sin(t/2)+cos(t/2)]} cos(t/2)/sin(t/2)=1/2 cot(t/2)=1/2 tan(t/2)=2 由万能公式：cost={1-[tan(t/2)]^2}/{1+[tan(t/2)]^2} =(1...全部

  (1+sinθ+cosθ)/(1+sinθ-cosθ)=1/2 解：[(1+cost)+sint]/[(1-cost)+sint]=1/2 左边=[2(cos(t/2))^2+2sin(t/2)cos(t/2)] /[2(sin(t/2))^2+2sin(t/2)cos(t/2)] =2cos(t/2)[cos(t/2)+sin(t/2)] /{2sin(t/2)[sin(t/2)+cos(t/2)]} cos(t/2)/sin(t/2)=1/2 cot(t/2)=1/2 tan(t/2)=2 由万能公式：cost={1-[tan(t/2)]^2}/{1+[tan(t/2)]^2} =(1-2^2)/(1+2^2) =-3/5。
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