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¶

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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.

Richard P. | Fairfax County Tutor for HS Math and ScienceFairfax County Tutor for HS Math and Sci...

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

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