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2015-04-24T17:49:04+05:30

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We use the identities
\cos(x-y)\cos(x+y)=\cos^2y-\sin^2x
\cos(3x)=4\cos^3(x)-3\cos(x)

Start with left hand side :
\cos(a)\cos(60-a)\cos(60+a)\\=\cos(a)\left(\cos^2(a)-\sin^2(60)\right)\\=\cos(a)\left(\cos^2(a)-\frac{3}{4}\right)\\=\dfrac{4\cos^3(a)-3\cos(a)}{4}\\=\dfrac{\cos(3a)}{4}
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can you derive the 2nd identity and thanks if you did
should be easy with angle sum identity : cos(3x) = cos(2x+x) = cos(2x)cos(x)-sin(2x)sin(x) = (2cos^2(x)-1)cos(x) - 2sin(x)cos(x)sin(x) = ...
2015-04-25T05:58:53+05:30

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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.
We know that standard identity:

     2 Cos A Cos B =  Cos (A+B) + Cos (A-B)

        here,    A = 60-a  and   B = 60+a

Cos (a)  Cos (60 - a )  Cos (60+a)
  = Cos (a) * 1/2 * [ Cos (60-a+60+a) + Cos (60-a- 60-a) ]
  = Cos (a) *  1/2 * [ Cos 120  + Cos (-2a)  ]
  = 1/2 * Cos (a) * [ -1/2 + Cos (2a) ]
  = -1/4 * Cos (a) + 1/2 Cos (a)  Cos (2a)       --- apply the above identity again
  = -1/4 * Cos (a)  + 1/2  * 1/2 [ Cos 3a + cos a]
  = 1/4 * Cos (3a)


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click on thank you button above pls