三角函數疑問求解

請問下面的式算的出來嗎?
－3Sin3xCosx+Cos3xSinx =?
Sol
Sin3x=3Sinx－4Sin^3 x
Cos3x=4Cos^3 x－3Cosx
－3Sin3xCosx+Cos3xSinx
=－3cosx(3Sinx－4Sin^3x)+Sinx(4Cos^3 x－3Cosx)
=－9SinxCosx+12Sin^3 xCosx+4SinxCos^3 x－3SinxCosx
=－12SinxCosx+12Sin^3 xCosx+4SinxCos^3 x
=4SinxCosx(－3+3Sin^2 x+Cos^2 x)
=4SinxCosx(－3+2Sin^2 x+1)
=4SinxCosx(－2+2Sin^2 x)
=4SinxCosx(－2Cos^2 x)
=－8SinxCos^3 x

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sinAcosB = (1/2)[sin(A+B) + sin(A-B)]
cosAsinB = (1/2)[sin(A+B) - sin(A-B)]
sin(2A) = 2sinAcosA
cos(2A) = 2(cosA)^2 - 1

-3sin(3x)cos(x) + cos(3x)sin(x)
= (-3/2)[sin(4x) + sin(2x)] + (1/2))[sin(4x) - sin(2x)
= -sin(4x) - 2sin(2x)
= -2sin(2x)cos(2x) - 2sin(2x)
= -2sin(2x)[cos(2x) + 1]
= -4sin(x)cos(x)[2(cos(x))^2 - 1 + 1]
= -8sin(x)[cos(x)]^3