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The Brainliest Answer!

### This Is a Certified Answer

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

**=Cos2xCosx - Sin2xSinx**

**=(2Cos**

**²**

**x - 1)Cosx - 2SinxCosxSinx**

**[ Cos2x = 2Cos²x -1 ; Sin2x = 2Sinx.Cosx ]**

**Now multiplying the cosx with the term in the bracket,**

**=2Cos**

**³**

**x - Cosx - 2Cosx.Sin**

**²**

**x**

**= 2Cos**

**³**

**x - Cosx (1 + 2Sin**

**²**

**x)**

**= 2Cos**

**³**

**x - Cosx (1 + 2 ( 1 - Cos**

**²**

**x) [ Sin**

**²x = 1-Cos²x]**

**= 2Cos**

**³**

**x - Cosx ( 1+ 2 - 2Cos**

**²**

**x)**

**Now adding the terms in the bracket,**

**=2Cos**

**³**

**x - Cosx (3 -2Cos**

**²**

**x)**

**=2Cos**

**³**

**x - 3Cosx + 2Cos**

**³**

**x**

**Now adding the like terms :**

**We get,**

**=4Cos**

**³**

**x - 3cosx (proved)**

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Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

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

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**