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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
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.

1696_13429_110108-154410.jpg

Sudheesh Singanamalla
114 Points
13 years ago

Thank you everyone

Kushagra Madhukar
askIITians Faculty 628 Points
3 years 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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