Respected Sir/Madam, Please help me with Q.8 given on the page in the attached image. Ans: is n=6



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rakesh05's profile pic

Posted on (Answer #1)

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`


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


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



`=(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.


aruv's profile pic

Posted on (Answer #2)

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










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



compairing (i) and (ii)

It will be  n=6

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