作业帮cos(270°-θ)cos^2(180°-θ)+cos^2(90°-θ)cos(-90°+θ)答案是-sinθ请问如何算出?
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/12 15:22:54
cos(270°-θ)cos^2(180°-θ)+cos^2(90°-θ)cos(-90°+θ)答案是-sinθ请问如何算出?
cos(270°-θ)cos^2(180°-θ)+cos^2(90°-θ)cos(-90°+θ)
答案是-sinθ
请问如何算出?
cos(270°-θ)cos^2(180°-θ)+cos^2(90°-θ)cos(-90°+θ)答案是-sinθ请问如何算出?
cos（270-θ）=cos（360-90-θ）=cos（-90-θ）=cos（90+θ）= -sinθ
cos^2(180-θ)=cos^2(θ)
cos^2(90-θ)=sin^2(θ)
cos(-90+θ)=sinθ
∴原式=-sinθcos^2(θ)+sinθsin^2(θ)
=sinθ(sin^2(θ)-cos^2(θ))
=-sinθcos2θ
额 算到这里不对了.我觉得题目应该打错了,左后一项应该是cos(90+θ)才对吧.
cos(270°-θ)cos^2(180°-θ)+cos^2(90°-θ)cos(-90°+θ)
=-sinθcos^2(θ)-sin^2(θ)sinθ
=-sinθ(cos^2(θ)+sin^2(θ))
=-sinθ
cos(270°-θ)cos^2(180°-θ)+cos^2(90°-θ)cos(-90°+θ)
=-sinθcos^2θ+sin^2θ(-sinθ)
=-sinθ(cos^2θ+sin^2θ)
=-sinθ
cos(270-θ)=cos(270-θ-360)=cos(-90-θ)=-sinθ
cos(270°-θ)cos²(180°-θ)+cos²(90°-θ)cos(-90°+θ)
＝sin(-θ)cos²(θ)+sin²(θ)sin(θ)
＝-sinθ[cos²θ-sin²θ]
＝-sinθcos2θ
答案错了吧