以知cos(a+β )=-1,求证sin(2a+β )+sinβ=0
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以知cos(a+β)=-1,求证sin(2a+β)+sinβ=0以知cos(a+β)=-1,求证sin(2a+β)+sinβ=0以知cos(a+β)=-1,求证sin(2a+β)+sinβ=0cos(
以知cos(a+β )=-1,求证sin(2a+β )+sinβ=0
以知cos(a+β )=-1,求证sin(2a+β )+sinβ=0
以知cos(a+β )=-1,求证sin(2a+β )+sinβ=0
cos(a+b)=-1
sin(a+b)=0
sin(2a+b)+sinb=2(sin(a+b)*cosa)=0
以知cos(a+β )=-1,求证sin(2a+β )+sinβ=0
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