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Given    `sin^3xsin3x=sum_(m=0)^n c_m cosmx`           (1)

We have to find the value of n.

Now      `sin^3xsin3x=(1/2)sin^2x.2sinxsin3x`

=`(1/2)(1-cos^2x)(cos2x-cos4x)`

As 2sinA sinB=cos(A-B)-cos(A+B)

`=(1/4)(2-2cos^2x)(cos2x-cos4x)`

As `2cos^2x-1=cos2x`

`=(1/4)(1-(2cos^2x-1))(cos2x-cos4x)`

`=(1/4)(1-cos2x)(cos2x-cos4x)`

`=(1/4)(cos2x-cos4x-cos^2 2x+cos2x cos4x)`

`=(1/4)(cos2x-cos4x-(1/2)(2cos^2 2x-1)-(1/2)+(1/2)(cos6x+cos2x))`

As 2cosAcosB=cos(A+B)+cos(A-B)

`=(-1/8)+(3/8)cos2x-(3/8)cos4x+(1/8)cos6x`         (2)

Comparing equation (1) and (2) we get highest value equivalent to `cosmx`  is  `cos6x` .

So the highest value of m is 6.

So            n=6.

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`sin(3x)=3 sin(x)-4sin^3(x)`

`4sin^3(x)=3sin(x)-sin(3x)`

`sin^3(x)=(1/4)(3sin(x)-sin(3x))`

`Thus`

`LHS=sin^3(x)sin(3x)`

`=(1/4)(3sin(x)-sin(3x))sin(3x)`

`=(3/4)sin(x)sin(3x)-(1/4)sin^2(3x)`

`=(3/8)(2sin(x)sin(3x))-(1/4)(1-cos(6x))/2`

`=(3/8)(cos(3x-x)-cos(3x+x))-1/8+(1/8)cos(6x)`

`=(3/8)cos(2x)-(3/8)cos(4x)-1/8+(1/8)cos(6x)`

`=-1/8+(3/8)cos(2x)-(3/8)cos(4x)+(1/8)cos(6x)`     (i)

`sum_(m=0)^nC_mcos(mx)=C_0+C_1cos(x)+C_2cos(2x)+....+C_ncos(nx)`

(ii)

compairing (i) and (ii)

It will be  n=6