己知sin^2θ=sinα cosα 求cos2θ=2sin(45°-α) cos(45°+α) 急!~~

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/09 10:27:19
己知sin^2θ=sinα cosα 求cos2θ=2sin(45°-α) cos(45°+α) 急!~~

己知sin^2θ=sinα cosα 求cos2θ=2sin(45°-α) cos(45°+α) 急!~~
己知sin^2θ=sinα cosα 求cos2θ=2sin(45°-α) cos(45°+α) 急!~~

己知sin^2θ=sinα cosα 求cos2θ=2sin(45°-α) cos(45°+α) 急!~~
证明:
∵ 2sin(45°-α) cos(45°+α)
=2sin(45°-α) cos[90°-(45°-α)]
=2sin(45°-α) sin(45°-α)
=2sin²(45°-α)
=1-cos(90°-2α)
=1-sin2α
=1-2sinαcosα
=1-2sin²θ
=cos2θ
∴ cos2θ=2sin(45°-α) cos(45°+α)