# evaluate cosx . cos2x . cos3x !!the answer is given ascosx.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] / 4please explain how this step has come from the previous step !thanks in advance

Amit Saxena
35 Points
13 years ago

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
37 Points
13 years ago

Dear Sudheesh

See The Pic for the solution.

Sudheesh Singanamalla
114 Points
13 years ago

Thank you everyone

3 years ago
Dear student,

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