已知3sinβ=sin(2a+β)求证tan(a+β)=2tan a

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已知3sinβ=sin(2a+β)求证tan(a+β)=2tana已知3sinβ=sin(2a+β)求证tan(a+β)=2tana已知3sinβ=sin(2a+β)求证tan(a+β)=2tana因

已知3sinβ=sin(2a+β)求证tan(a+β)=2tan a
已知3sinβ=sin(2a+β)
求证tan(a+β)=2tan a

已知3sinβ=sin(2a+β)求证tan(a+β)=2tan a
因为3sinβ=sin(2a+β),推出3sin(a+β-a)=sin(a+β+a)
展开3sin(a+β)cosa-3cos(a+β)sina=sin(a+β)cosa+cos(a+β)sina
合并2sin(a+β)cosa=4cos(a+β)sina→sin(a+β)cosa=2cos(a+β)sina
两边同时除cos(a+β)cosa得tan(a+β)=2tan a