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

Sudheesh Singanamalla Grade: 12

        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

8 years ago

Answers : (3)

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Amit Saxena

35 Points

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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8 years ago

Approved

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