Question

prove that... cos(a)cos(60-a)cos(60+a)=(cos(3a))/4

Answer 1

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 :
$\cos(a)\cos(60-a)\cos(60+a)\\=\cos(a)\left(\cos^2(a)-\sin^2(60)\right)\\=\cos(a)\left(\cos^2(a)-\frac{3}{4}\right)\\=\dfrac{4\cos^3(a)-3\cos(a)}{4}\\=\dfrac{\cos(3a)}{4}$

Answer 2

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)