2014-07-28T19:37:02+05:30

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cos3x can be written as cos(2x+ x)  [ cos (a+b) = Cosa.Cosb -Sina.Sinb]

=Cos2xCosx - Sin2xSinx
=(2Cos²x - 1)Cosx - 2SinxCosxSinx [ Cos2x = 2Cos²x -1 ; Sin2x = 2Sinx.Cosx ]
Now multiplying the cosx with the term in the bracket,
=2Cos³x - Cosx - 2Cosx.Sin²x
= 2Cos³x - Cosx (1 + 2Sin²x)
= 2Cos³x - Cosx (1 + 2 ( 1 - Cos²x)  [ Sin²x = 1-Cos²x]
= 2Cos³x - Cosx ( 1+ 2 - 2Cos²x)
Now adding the terms in the bracket,
=2Cos³x - Cosx (3 -2Cos²x)
=2Cos³x - 3Cosx + 2Cos³
Now adding the like terms :
We get,
=4Cos³x - 3cosx (proved)
2014-07-28T21:07:25+05:30

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Use Cos A+B  = cos A cos B - SIN A sin B

Cos 3x  = cos (2x+x) = cos 2x  cos x - sin 2x   sin x
= (2 cos² x - 1)  cos x    -  2 sinx cos x   sinx
= 2 cos³ x  - cos x - 2 sin² x cos x
= 2 cos³ x - cos x - 2 (1 - cos² x ) cos x
= 2 cos² x - cos x - 2 cos x + 2 cos ³ x
= 4 cos³ x - 3 cos x
=========================================
From  R H S to L H S
use 2  cos² x - 1  = cos 2x

R H  S  =   cos x  (4 cos² x - 2  -  1)  = cos x  ( 2 cos 2x - 1 )
= 2 cos x cos 2x  -  cos x  = [ cos 3x + cos x ] - cos x
= cos  x  = L H S
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