Does \tan\frac{\theta}{2} equal \frac{\sin\theta}{1+\cos\theta}?

[bob1182006]

This isn't a homework question, it's just for personal enrichment.

I've been trying to prove that \tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}

I tried starting off with \tan\frac{\theta}{2}=\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}

is this even the right way to start the proof? if so could someone point me the right way please? I'd really appreciate it.

 

[courtrigrad]

note that \tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta} is equivalent to:

\tan\frac{\theta}{2}=\frac{1-\cos \theta}{\sin \theta}

We know that \frac{1-\cos 2\theta}{\sin 2\theta} = \tan \theta

Can you go from there?

 

[bob1182006]

ah I see it now hehe thank you verry much.

 

[murshid_islam]

\frac{\sin\theta}{1+\cos\theta} = \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}} = \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} = \tan\frac{\theta}{2}