What is the Origin of the Trigonometric Identity?

[Piano man]

I'm reading through a solution to a problem and at one point the following identity is used:

\frac{1-\cos(\beta)}{\sin(\beta)}=\frac{\sin(\frac{\beta}{2})}{\cos(\frac{\beta}{2})}

I've been trying to figure out where this comes from but with haven't got it yet.
Any ideas?

 

[AlephZero]

Just substitute in the well known formulas

1 = cos^2 b/2 + sin^2 b/2
cos b = cos^2 b/2 - sin^2 b/2
sin b = 2 cos b/2 sin b/2

 

[sadaf2605]

Sin (a + a) = sin a cos a+ cos a sin a
=> Sin 2a = 2Sin a Cos a
=> SIN B = 2 Sin (B/2) Cos (B/2) [2a = B] . . . . (i)

Cos (a + a) = cos a cos a - sin a sin b
=> cos 2a = cos^2 (a) - sin^2 (a)
=> cos 2a = (1 - sin^2 (a) ) - sin^2 a
=> cos 2a = 1 - 2sin^2 (a)
=> 1 - cos 2a = 2s n^2 (a)
=> 1 - cos B = 2sin^2 (B/2) . . . (ii)

n0w (ii) / (i) you should do it n0w urself . . . Am en0ugh tired already

 

[Bohrok]

You could start with the half-angle identities for sin and cos:

And since tanx = sinx/cosx, dividing them gives you \pm\sqrt{\frac{1 - cos\theta}{1 + \cos\theta}}
which can be simplified to \frac{\sin\theta}{1 + \cos\theta} = \frac{1 - \cos\theta}{\sin\theta}
(the ± goes away because each of the last two expressions give you the correct sign of tan^θ/₂)

You could also start with

and substitute x = 2θ → θ = x/2, then solve for tan^x/₂ which will end up giving you the same expressions above.

 

[TylerH]

No need to use cos and sin half angle formulas. sin over cos is tan, so use tan. https://secure.wikimedia.org/wikipedia/en/wiki/Tangent_half-angle_formula