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## Prove that 1 + sin b - cos b =2sin1/2b cos1/2b +sin1/2b

Prove that 1 +sin b -cos b=2sin 1 b cos 1 b +sin 1 b
2          2           2

and hence solve 1 +sin b -cos b =0  0 = <b=<2¶

You forgot the parentheses:

1 + sin b - cos b = 2sin b/2 (cos b/2 +sin b/2)

To prove this identity, use the double angle formulas

sin b = 2 sin b/2 cos b/2
cos b = 1 - 2 sin² b/2

Then

1 + sin b - cos b = 1 + 2 sin b/2 cos b/2 - (1 - 2 sin² b/2)
= 2 sin b/2 cos b/2 + 2 sin² b/2 = 2 sin b/2 (cos b/2 + sin b/2 )

Now we can solve 1 +sin b -cos b =0

2 sin b/2 (cos b/2 + sin b/2 )=0

Case I: sin b/2 =0.
b/2 = n*Pi,  b =2n*Pi, where n is any integer
Case II: cos b/2 + sin b/2 =0.
tan b/2 = -1,
b/2 = -Pi/4 + m*Pi,  b = -Pi/2 + 2m*Pi, where m is any integer.

If 0≤b≤2, the only solution is b=0.
The term 2 sin b/2 cos b/2   =   sin b          (double angle formula)

This leaves     1 -cos b = sin b/2        This is not true for all values of b.     Squaring both side and
using the half angle formula   for sin b/2   results in

(1-cos b)2 = (1- cos b)/2     which implies 1-cos b = 1/2

So b =  pi/3  or   5 pi/3