The value of (sin8x+7sin6x+18sin4x+12sin2x)/(sin7x+6sin5x+12sin3x) is equal toa.2cosxb.cosxc.2sinxd.sinx

Hi Menka,

The first thing that should come to your mind on seeing sum of angle of sin's or sum of angle of cos's is the identity:

2SinACosB = Sin(A+B) + Sin(A-B)

or the identity 2SinASinB = Cos(A-B) - Cos(A+B)

So, consider applying the first identity in the Dr, as all terms in the Nr as sum of Sin's

So sin7x+6sin5x+12sin3x = [2sinxsin7x+2sinx(6sin5x)+2sinx(12sin3x)]/2sinx

So the terms in the square bracket, we apply the first identity, and easily see that it is equal to sin8x+7sin6x+18sin4x+12sin2x, which is the Nr in the question.

So the given fraction is equal to 2sinx.... Option (C).

Best Regards,

Ashwin (IIT Madras).

Hi Menka,

A very silly mistake, in the answer....

As we are applying the first identity, we use 2Cosx.... and the identity of 2SinACosB (as mentioned above)....

Kindly note the answer would be Option (B).

Moderators kindly approve both replies, as this one is a continuation of the previous one.

Best Regards,

Ashwin (IIT Madras).