evaluate cosx . cos2x . cos3x !!

the answer is given as

cosx.cos2x.cos3x = 2cosxxos2x*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

Question

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

Amit Saxena · Answer

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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Mayank Jain · Answer

Dear Sudheesh See The Pic for the solution.

Sudheesh Singanamalla · Answer

Thank you everyone

Kushagra Madhukar · Answer

Dear student, Please find the solution to your problem below. cosx . cos2x . cos3x = ½ (2cosx . cos3x) . cos2x = ½ (cos4x + cos2x) . cos2x = ½ (cos4x . cos2x + cos 2 2x ) = ¼ (2cos4x . cos2x + 2cos 2 2x) = ¼ ({cos6x + cos2x} + {cos4x + 1}) = (1 + cos6x + cos4x + cos2x) / 4 Hope it helps. Thanks and regards, Kushagra

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evaluate cosx . cos2x . cos3x !! the answer is given as cosx.cos2x.cos3x = 2cosxxos2x*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

evaluate cosx . cos2x . cos3x !!

the answer is given as

cosx.cos2x.cos3x = 2cosxxos2x*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

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