.已知sina+sinb+siny=0,cosa+cosb+cosy=0,求cos(b-y)的值.(解题
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.已知sina+sinb+siny=0,cosa+cosb+cosy=0,求cos(b-y)的值.(解题
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.已知sina+sinb+siny=0,cosa+cosb+cosy=0,求cos(b-y)的值.(解题
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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