# Prove the identity: (cos x + cos y)^2 + (sin x – sin y)^2 = 2 + 2cos(x + y)

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### 2 Answers

The identity `(cos x + cos y)^2 + (sin x – sin y)^2 = 2 + 2cos(x + y)` has to be proved.

Start from the left hand side.

`(cos x + cos y)^2 + (sin x – sin y)^2`

=> `cos^2 x` + `cos^2 y` + `2*cos x*cos y` + `sin^2 x` + `sin^2 y` - `2*sin x*sin y`

=> `cos^2 x + sin^2 x` + `cos^2 y + sin^2 y` + `2*cos x*cos y - 2*sin x*sin y`

=> `1 + 1 + 2*(cos x*cos y - sin x*sin y)`

Use the relation `cos(x + y) = cos(x)cos(y) - sin(x)sin(y)`

=> 2 + 2*cos (x + y)

**This proves that (cos x + cos y)^2 + (sin x – sin y)^2 = 2 + 2cos(x + y)**

`(cosx+cosy)^2+(sinx-siny)^2=(2cos((x+y)/2)cos((x-y)/2))^2+(2sin((x-y)/2)cos((x+y)/2))^2=`

`=4cos^2((x+y)/2)cos^2((x-y)/2)+4sin^2((x-y)/2)cos^2((x+y)/2)=`

`=4cos^2((x+y)/2)(cos^2((x-y)/2)+sin^2((x-y)/2))=`

`=4cos^2((x+y)/2)` `=2(1+cos(x+y))=2+2cos(x+y)`