已知sinθ+sin∧2(θ)=1,求3cos∧2(θ)+cos∧4(θ)-2sinθ
问题描述：
已知sinθ+sin∧2(θ)=1,求3cos∧2(θ)+cos∧4(θ)-2sinθ
答
sinθ+sin∧2(θ)=1,
sinθ=1-sin∧2(θ),
3cos∧2(θ)+cos∧4(θ)-2sinθ
=3(1-sin^2(θ))+((1-sin^2(θ)))^2-2sinθ
=3sinθ+(sinθ)^2-2sinθ
=sinθ+(sinθ)^2
=1
答
sinθ+sin^2θ=1
sinθ=1-sin^2θ=cos^2θ
3cos^2θ+cos^4θ-2sinθ
=3sinθ+sin^2θ-2sinθ
=sin^2θ+sinθ
=1