Precalculus-Trigonometry

Question

tan 1/2 b = 1-cos b/sin b changed to

tan b = 1-sin 2b/sin 2b

Why the 1 in numerator 1-cos b is not changed to 2 after calculating

Answer 1

Assuming Dayv O.'s reading as

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

let x = b / 2

tan x = (1 - cos 2x ) / sin 2x

tan x = (1 - cos² x + sin² x) / (2 sin x cos x)

tan x = (2 sin² x) / (2 sin x cos x)

tan x = sin x / cos x

Examining the second equation

tan b = (1 - sin 2b) / sin 2b

tan b = [1 - (2 sin b cos b)] / (2 sin b cos b)

tan b = [1 / (2 sin b cos b)] - 1

The derivation has not been demonstrated. The post should be reviewed for accuracy.

Answer 2

It can be proved tan(b/2)=(1-cos(b))/sin(b)

let x=b/2

have tan(x)=(1-cos(2x))/sin(2x)

tan(x)=(1-cos²(x)+sin²(x))2sin(x)cos(x)

tan(x)=(2sin²(x))2sin(x)cos(x)=tan(x)

does that clear up why 1 in numerator never became 2?