已知sinθ+sin²θ=1,求3cos²θ+cos^4θ-2sinθ+1的值

Question

如题所述

Answer 1

sinθ+sin²θ=1 ，sinθ=1-sin²θ ，sinθ=cos²θ
代入右式
3cos²θ+cos^4θ-2sinθ+1
=3cos²θ+cos^4θ-2cos²θ+1
=cos²θ+cos^4θ+1
又sinθ=cos²θ ，sin²θ=cos^4θ ，1-cos²θ=cos^4θ ，cos²θ+cos^4θ=1
代入
cos²θ+cos^4θ+1
=1+1
=2

Answer 2

解：由sinθ+sin²θ=1及cos²θ+sin²θ=1得：
cos²θ=sinθ
3cos²θ+cos^4θ-2sinθ+1
=3sinθ+sin²θ-2sinθ+1
=1+sinθ+sin²θ
=2

Answer 3

因为 sinθ+sin²θ=1
有sinθ=1 - sin²θ =cos²θ , sin²θ == cos^4²θ
所以，
3cos²θ+cos^4θ-2sinθ+1
= 3cos²θ + sin²θ - 2cos²θ +1
= cos²θ + sin²θ +1
= 1 + 1
= 2