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 [itex]\tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}[/itex]

I tried starting off with [tex]\tan\frac{\theta}{2}=\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}[/tex]

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 [tex]\tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}[/tex] is equivalent to:

[tex]\tan\frac{\theta}{2}=\frac{1-\cos \theta}{\sin \theta}[/tex]

We know that [tex]\frac{1-\cos 2\theta}{\sin 2\theta} = \tan \theta[/tex]

Can you go from there?

 

[bob1182006]

ah I see it now hehe thank you verry much.

 

[murshid_islam]

[tex]\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}[/tex]