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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

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

Grade:12

4 Answers

Amit Saxena
35 Points
10 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
10 years ago

Dear Sudheesh

See The Pic for the solution.

1696_13429_110108-154410.jpg

Sudheesh Singanamalla
114 Points
10 years ago

Thank you everyone

Kushagra Madhukar
askIITians Faculty 629 Points
one year ago
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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