已知sina=sinb=siny=0,cosa=cosb=cosy=0,求cos(b-y)的值
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已知sina=sinb=siny=0,cosa=cosb=cosy=0,求cos(b-y)的值
cos(b-y)=cosb*cosy+sinb*siny=0*0+0*0=0
如果问题是sina+sinb+siny=0 (1),cosa+cosb+cosy=0 (2)
那么 (sinb+siny)^2=(sina)^2 (3)
(cosb+cosy)^2=(cosa)^2 (4)
相加(3) (4) 式 可得
(sinb)^2+(cosb)^2+(siny)^2+(cosy)^2+2*(sinb*siny+cosb*cosy)=(sina)^2+(cosa)^2
即为 1+1+2cos(b-y)=1
故 cos(b-y)=-0.5
如果问题是sina+sinb+siny=0 (1),cosa+cosb+cosy=0 (2)
那么 (sinb+siny)^2=(sina)^2 (3)
(cosb+cosy)^2=(cosa)^2 (4)
相加(3) (4) 式 可得
(sinb)^2+(cosb)^2+(siny)^2+(cosy)^2+2*(sinb*siny+cosb*cosy)=(sina)^2+(cosa)^2
即为 1+1+2cos(b-y)=1
故 cos(b-y)=-0.5
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