Question

Prove that:

cosa + cosb + cosc + cos (a +b +c) = 4 cos(a+b/2).cos (b+c/2).cos (c+a/2)​

Answer 1

Step-by-step explanation:

Let, a= 'α', b= 'β' & c= 'γ'

we have,

$\cos( \alpha ) + \cos( \beta ) + \cos( \gamma ) + \cos( \alpha + \beta + \gamma )$

Applying C & D formulas,

$=2 \cos( \frac{ \alpha + \beta }{2} ) \cos( \frac{ \alpha - \beta}{2} ) + 2 \cos( \frac{ \alpha + \beta }{2} + \gamma ) \cos( \frac{ \alpha + \beta }{2} )$

$= 2 \cos( \frac{ \alpha + \beta }{2} ) ( \cos( \frac{ \alpha - \beta }{2} ) + \cos( \frac{ \alpha + \beta }{2} + \gamma ) )$

$= 2 \cos( \frac{ \alpha + \beta }{2} ) (2 \cos( \frac{ \alpha + \gamma }{2} ) \cos( \frac{\beta + \gamma}{2} ) )$

putting α= a, β= b, γ=c,

$= 4 \cos( \frac{a + b}{2} ) \cos( \frac{b + c}{2} ) \cos( \frac{c + a}{2} )$