已知sina+sinβ+siny=0,cosa+cosβ+cosy=0,求cos(β-y)的值?
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已知sina+sinβ+siny=0,cosa+cosβ+cosy=0,求cos(β-y)的值?已知sina+sinβ+siny=0,cosa+cosβ+cosy=0,求cos(β-y)的值?已知si
已知sina+sinβ+siny=0,cosa+cosβ+cosy=0,求cos(β-y)的值?
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sinb+siny=-sina
cosb+cosy=-cosa
分别平方并相加
且利用恒等式sin²x+cos²x=1
所以2+2(cosbcosy+sinbsiny)=1
所以cos(b-y)=cosbcosy+sinbsiny=-1/2
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