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

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2 Answers

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
 
Let b be in radians.
 
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