Prove that `cos (x/2)=sqrt ((1/2)(1 + cosx))`

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The identity `cos (x/2)=sqrt ((1/2)(1 + cosx))` has to be proved.

`cos (x/2)=sqrt ((1/2)(1 + cosx))`

let `x/2= y` , this gives the identity to be proved as `cos y = sqrt((1 + cos 2y)/2)`

`sqrt((1 + cos 2y)/2)`

=> `sqrt((1 + 2*cos^2y -1 )/2)`

=> `sqrt((2*cos^2y)/2)`

=> `sqrt(cos^2y)`

=> `cos y`

This proves `cos y = sqrt((1 + cos 2y)/2)` .

** If `x/2` is substituted for y it proves: `cos (x/2)=sqrt ((1/2)(1 + cosx))` **

`cos(x/2)=sqrt((cosx+1)/2)`

squaring boht sides:

`cos^2(x/2)=(cosx+1)/2=(cos^2(x/2)-sen^2(x/2)+1)/2=`

`cos^2(x/2)=(cos^2(x/2)+cos^2(x/2))/2=` `(2cos^2(x/2))/2=cos^2(x/2)`

The general equatio is shown as this:

cos(a+b) = cosa*cosb - sina*sinb

when a = b = x/2,

cosx = cos(x/2)*cos(x/2) - sin(x/2)*sin(x/2)

= (cos(x/2))^2 - (sin(x/2))^2

= (cos(x/2))^2 - (1-cos(x/2)^2) ((siny)^2 + (cosy)^2 = 1)

= 2cos(x/2)^2 - 1

cosx = 2(cos(x/2))^2 - 1

=> 2(cos(x/2))^2 = cosx + 1

=> (cos(x/2))^2 = (cosx+1)/2

Therefore, cos(x/2) = ((cosx + 1)/2)^(1/2)

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