若(sinθ)^6+(cosθ)^6=13/16,则sin2θ=

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若(sinθ)^6+(cosθ)^6=13/16,则sin2θ=
若(sinθ)^6+(cosθ)^6=13/16,则sin2θ=

若(sinθ)^6+(cosθ)^6=13/16,则sin2θ=
(sinθ)^6+(cosθ)^6=13/16
[(sinθ)^2+(cosθ)^2][(sinθ)^4-(sinθ)^2(cosθ)^2+(cosθ)^4]=13/16
(sinθ)^4-(sinθ)^2(cosθ)^2+(cosθ)^4=13/16
[(sinθ)^2+(cosθ)^2]^2-3(sinθ)^2(cosθ)^2=13/16
1-3(sinθcosθ)^2=13/16
3(sinθcosθ)^2=3/16
(sinθcosθ)^2=1/16
sin2θ^2=1/4
sin2θ=±1/2

若sinθ+sin^2θ=1,则cos^2θ+cos^4θ+cos^6θ 已知cosθ+cosθ^2=1,则sinθ^2+sinθ^6+sinθ^8= 已知cosθ+cosθ=1,则sinθ+sin∧6θ+sin∧8θ= (1-sin^6θ-cos^6θ)/(1-sin^4θ-cos^4θ) 若sinθ-cosθ=1/5,则cosθ乘以sinθ= 若tanθ=1/2则(cosθ+sinθ/cosθ-sinθ),1-sinθcosθ+2cos^2θ= 若sinα+cosα=1,求证:(sinα)^6+(cosα)^6=1 若sinα+sin^2α=1,则cos^2α+cos^6α+cos^8α=? 若sinα+sin^2α=1,则cos^2α+cos^4α+cos^6α=? 设cosθ+cos^2θ=1,则sin^2θ+sin^6θ+sin^8θ的值为求证sinθ/(1+cosθ)+(1+cosθ)/sinθ=2/sinθ 已知(4sinθ-2cosθ)/(3sinθ+5cosθ)=6/11,求5cos^2θ/(sin^2θ+2sinθcosθ-3cos^2θ) 若sinθ-2cosθ=0,求：（1）sin^2θ+sinθcosθ-2cos^θ的值（2）1/cos^2θ+2sinθcosθ的值 2sinθ-cosθ=1 (sinθ+cosθ+1)/(sinθ-cosθ+1)如题已知2sinθ-cosθ=1 求(sinθ+cosθ+1)/(sinθ-cosθ+1) f(θ)=【sinθcosθ/(sinθ+cosθ+1)】+sinθcosθ化简若sin θ-cos θ 分之sin θ+cos θ=2 则sin θcos θ 是若(sinθ+cosθ)/(sinθ-cosθ)=2,则sinθcosθ的值是求证：(1+sinθ-cosθ)/(1+sinθ+cosθ)=sinθ/（cosθ+1）