2[√3/2sin(wx+φ)-1/2cos(wx+φ)] =2sin(wx+φ-π/6)这一步怎么得得、?说下运用什么知识
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2[√3/2sin(wx+φ)-1/2cos(wx+φ)] =2sin(wx+φ-π/6)这一步怎么得得、?说下运用什么知识
2[√3/2sin(wx+φ)-1/2cos(wx+φ)] =2sin(wx+φ-π/6)
这一步怎么得得、?说下运用什么知识
2[√3/2sin(wx+φ)-1/2cos(wx+φ)] =2sin(wx+φ-π/6)这一步怎么得得、?说下运用什么知识
consider: sin(A-B)= sinAcosB-cosAsinB
2[(√3/2)sin(wx+φ)-(1/2)cos(wx+φ)]
=2[sin(wx+φ)cos(π/6)-cos(wx+φ)sin(π/6)]
=2sin(wx+φ-π/6)
cosˆ2wx+2√3sinwxcoswx-sinˆ2wx+1怎么化简啊
已知函数f x=√3sin(wx+φ/2)*cos(wx+φ/2)+sin^2(wx+φ/2)(w>0,0
2[√3/2sin(wx+φ)-1/2cos(wx+φ)] =2sin(wx+φ-π/6)这一步怎么得得、?说下运用什么知识
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