已知COS(A+B)=0 求证SIN(A+2B)=SINA
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已知COS(A+B)=0 求证SIN(A+2B)=SINA
求...
求...
证明:由COS(A+B)=0,有:
COS(A+B)=cosAcosB-sinAsinB=0
所以cosAcosB=sinAsinB
SIN(A+2B)=sinAcos2B+sin2BcosA
=sinA[(cosB)^2-(sinB)^2]+2sinBcosBcosA
=sinA[(cosB)^2-(sinB)^2]+2sinA(sinB)^2
=sinA[(cosB)^2-(sinB)^2+2(sinB)^2]
=sinA[(cosB)^2+(sinB)^2]
=sinA
COS(A+B)=cosAcosB-sinAsinB=0
所以cosAcosB=sinAsinB
SIN(A+2B)=sinAcos2B+sin2BcosA
=sinA[(cosB)^2-(sinB)^2]+2sinBcosBcosA
=sinA[(cosB)^2-(sinB)^2]+2sinA(sinB)^2
=sinA[(cosB)^2-(sinB)^2+2(sinB)^2]
=sinA[(cosB)^2+(sinB)^2]
=sinA
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