# Prove that 1 + sin x - cos x = 2 sin(x/2)[cos(x/2) + sin(x/2)].

You need to open the brackets to the right side such that:

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

Notice that you may substitute `sin 2*(x/2)`  for `2 sin(x/2)cos(x/2)`  such that:

`1 + sin x - cos x =sin 2*(x/2) + 2sin^2(x/2)`

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

Reducing sin x both sides yields:

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

Interchanging the terms yields:

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

You may write `cos x = cos (x/2 + x/2)`  such that:

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

Using the formula `cos (a+b) = cos a*cos b - sin a*sin b`  yields:

`cos (x/2 + x/2) = cos(x/2)*cos(x/2) - sin(x/2)sin(x/2)`

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

Using the fundamental formula of trigonometry yields:

`cos^2(x/2) + sin^2(x/2) = 1 => cos^2(x/2) = 1 - sin^2(x/2)`

Substituting `1 - sin^2(x/2)`  for `cos^2(x/2` ) yields:

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

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

Hence, using the substitutions above yields:

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

Hence, since both sides become equal, the given identity `1 + sin x - cos x = 2 sin(x/2)[cos(x/2) + sin(x/2)]`  holds.

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Using the double-angle formulae

`sin2y = 2sin(y)cos(y)`,  `cos2y = cos^2y - sin^2y`

and letting `y = x/2`

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

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

(using `sin^2y +cos^y = 1`)

`therefore sinx - cosx + 1 = 2sin(x/2)cos(x/2) +2sin^2x/2`

`= 2sin(x/2)(cos(x/2) + sin(x/2))`  proof complete

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