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三倍角公式推导 tan3=sin3/cos3 =(sin2cos+cos2sin)/(cos2cos-sin2sin) =(2sincos^2()+cos^2()sin-sin^3())/(cos^3()-cossin^2()-2sin^2()cos) 上下同除以cos^3()，得： tan3=(3tan-tan^3())/(1-3tan^2()) sin3=sin(2+)=sin2cos+cos2sin =2sincos^2()+(1-2sin^2())sin =2sin-2sin^3()+sin-2sin^3()=3sin-4sin^3() cos3=cos(2+)=cos2cos-sin2sin =(2cos^2()-1)cos-2cossin^2() =2cos^3()-cos+(2cos-2cos^3()) =4cos^3()-3cos 即 sin3=3sin-4sin^3() cos3=4cos^3()-3cos | 
