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evaluate cosx . cos2x . cos3x !! the answer is given as cosx.cos2x.cos3x = 2*cosx*xos2x*cos3x/2 and so on until we get [2 cos^2 3x + 2 cos 3x.cos x]/4 which is furthur equal to [1+cos6x + cos4x + cos2x] / 4 please explain how this step has come from the previous step ! thanks in advance
Dear Sudeesh,
[2 cos^2 3x + 2 cos 3x.cos x]/4 :
you can use Cos (A+B) + Cos(A-B) for the last two terms and Cos 2A = 2Cos2A - 1 for the first term, in result you will get as :
[1+cos6x + cos4x + cos2x] / 4
All the best.
AKhilesh Shukla
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Dear Sudheesh
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Dear student,Please find the solution to your problem below. cosx . cos2x . cos3x = ½ (2cosx . cos3x) . cos2x= ½ (cos4x + cos2x) . cos2x= ½ (cos4x . cos2x + cos22x )= ¼ (2cos4x . cos2x + 2cos22x)= ¼ ({cos6x + cos2x} + {cos4x + 1})= (1 + cos6x + cos4x + cos2x) / 4 Hope it helps.Thanks and regards,Kushagra
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