Question

If `y=2^xsin^2x cosx` find `y_n`

Answer 1

`2^x=e(xlog2)=e(ax)` where a = log2

`(2sin^2x cosx)/2=(sin^1x cosx.sinx xx2)/2= (sinx.sin2x)/2 = (2sinx.sin2x)/(2xx2)=cosx/4-(cos3x)/4`

`therefore sin^2x cosx=(cosx)/4-(cos3x)/4`

`Y=(e^(ax)cosx)/4-(e^(ax)cos3x)/4`

`y_n=1//4 r^n ""_1 e^(ax) cos(x+"n"varphi_1)-1//4 r^n""_2e^(ax) cos(3x+n varphi_2)`

`y_n=1//4r^n""_1 2^(1x)cos(x+nvarphi_1)-1//4r^n""_2 2^(1x)cos(3x+nvarphi_2)`

`r_1=sqrt((log2)^2)+1`      `r_2=sqrt((log2)^2)+3^2`

`varphi_1 = tan^(-1)[1/(log2)]`        `varphi_2=tan^(-1)[3/(1og2)]`