求证sin^2α+sin^2β-sin^2αsin^2β+cos^2cos^2β=1
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/11 17:02:32
求证sin^2α+sin^2β-sin^2αsin^2β+cos^2cos^2β=1
求证sin^2α+sin^2β-sin^2αsin^2β+cos^2cos^2β=1
求证sin^2α+sin^2β-sin^2αsin^2β+cos^2cos^2β=1
左边=sin^2α(1-sin^2β)+sin^2β+cos^2cos^2β
=sin^2αcos^2β+cos^2cos^2β+sin^2β
=(sin^2α+cos^2α)cos^2β+sin^2β
=cos^2β+sin^2β
=1=右边
命题得证
求证:sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα
求证sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα
求证sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα
求证sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα
求证:sin²α+sin²β+2sinαsinβcos(α+β)=sin²(α+β)
求证:sinα+sinβ=2sin(α+β)/2 *cos(α-β)/2
求证sinα-sinβ=2cos(α+β)/2sin(α-β)/2
求证 sinαcosβ=1/2[sin(α+β)+sin(α-β)]
求证:[sin(2α+β)/2sinα]-cos(α+β)=sinβ/2sinα
求证:sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα
求证sin^2α+sin^2β-sin^2αsin^2β+cos^2cos^2β=1
已知sin(2α+ β)=5sinβ,求证:2sin(α+ β)=3sinα
求证:sin(α+β)cosα-1╱2sin(2α+β)=1╱2sinβ
求证!sinα+sinβ=2[sin(α+β)/2] [cos(α-β)/2]
已知sin(2α+β)=5sinβ,求证:2sin(α+β)=3tanα
已知sin(α+β)=1,求证sin(2α+β)+sin(2α+3β)=0
已知sin(α+β)=1,求证sin(2α+β)+sin(2α+3β)=0
已知sin(α+β)=1,求证:sin(2α+β)+sin(2α+3β)=0