Question

(cos alpha + cos beta)^2 +(sin alpha +sin beta)^2 =4cos^2(alpha -beta/2)

Answer 1

$\green{\large\underline{\sf{Solution-}}}$

Consider,

$\rm :\longmapsto\: {(cos\alpha + cos\beta )}^{2} + {(sin\alpha + sin\beta )}^{2}$

We know,

$\red{ \boxed{ \sf{ \:cosx + cosy = 2cos\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}}}$

and

$\red{ \boxed{ \sf{ \:sinx + siny = 2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}}}$

Let assume that x = alpha and y = beta

Using this result, we get

$\rm ={\bigg(2cos\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]\bigg) }^{2} + {\bigg(2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]\bigg) }^{2}$

$\rm = 4 {cos}^{2}\bigg[\dfrac{x + y}{2} \bigg] {cos}^{2}\bigg[\dfrac{x - y}{2} \bigg] + 4 {sin}^{2}\bigg[\dfrac{x + y}{2} \bigg] {cos}^{2}\bigg[\dfrac{x - y}{2} \bigg]$

$\rm = 4 {cos}^{2}\bigg[\dfrac{x - y}{2} \bigg] \bigg({cos}^{2}\bigg[\dfrac{x + y}{2} \bigg] + {sin}^{2}\bigg[\dfrac{x + y}{2} \bigg]\bigg)$

We know,

$\red{ \boxed{ \sf{ \: {sin}^{2}x + {cos}^{2}x = 1}}}$

So, using this, we get

$\rm \:  =  \: {4cos}^{2}\bigg[\dfrac{x - y}{2} \bigg] \times 1$

$\rm \:  =  \: {4cos}^{2}\bigg[\dfrac{x - y}{2} \bigg]$

Hence, Proved

Additional Information :-

$\red{ \boxed{ \sf{ \:2sinxcosy = sin(x + y) + sin(x - y)}}}$

$\red{ \boxed{ \sf{ \:2cosxcosy = cos(x + y) + cos(x - y)}}}$

$\red{ \boxed{ \sf{ \:2sinxsiny = cos(x - y) - cos(x + y)}}}$

$\red{ \boxed{ \sf{ \:cosx - cosy = 2sin\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{y - x}{2} \bigg]}}}$