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a,b为锐角,3sina+3sinb=1 ,3sin(2a)-2sin(2b)=0 求sin(a+2b)
人气:102 ℃ 时间:2020-06-02 08:40:58
解答
明如下由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
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