sinX+sinY+sinZ=0 同时满足cosX+cosY+cosZ=0求cos(X-Y)为何值?所有的条件就这些了,我只要答案,
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sinX+sinY+sinZ=0 同时满足cosX+cosY+cosZ=0求cos(X-Y)为何值?所有的条件就这些了,我只要答案,
sinX+sinY+sinZ=0 同时满足cosX+cosY+cosZ=0
求cos(X-Y)为何值?
所有的条件就这些了,
我只要答案,
sinX+sinY+sinZ=0 同时满足cosX+cosY+cosZ=0求cos(X-Y)为何值?所有的条件就这些了,我只要答案,
sinX + sinY = -sinZ ==>sinX^2+sinY^2+2sinXsinY=sinZ^2
cosX + cosY = -cosZ ==>cosX^2+cosY^2+2cosXcosY=cosZ^2
两式相加得:
sinXsinY+cosXcosY=-1/2=cos(X-Y)
(sinX+sinY)^2+(cosX+cosY)^2=(-sinz)^2+(-cosz)^2=1
左边=2+2cos(x-y)
所以所求为-1/2
sinx+siny+sinz=0;cosx+cosy+cosz=0;求cos(x-y)
sinX+sinY+sinZ=0 同时满足cosX+cosY+cosZ=0求cos(X-Y)为何值?所有的条件就这些了,我只要答案,
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