[SOLVED] How can i show this, trig identity

**maths900** · February 15th 2009, 05:53 AM

$\frac{sin\theta}{1-cos\theta}=\frac{1}{tan(\theta/2)}$

**Moo** · February 15th 2009, 05:58 AM

Hello,

Originally Posted by maths900

$\frac{sin\theta}{1-cos\theta}=\frac{1}{tan(\theta/2)}$

Do you know about Weierstrass substitution ?

Let $t=\tan (\theta/2)$

Then $\sin(\theta)=\frac{2t}{1+t^2}$ and $\cos(\theta)=\frac{1-t^2}{1+t^2}$

Otherwise... :
$\frac{1}{\tan(\theta/2)}=\frac{1}{\frac{\sin(\theta/2)}{\cos(\theta/2)}}=\frac{\cos(\theta/2)}{\sin(\theta/2)}$

Now the RHS :
$\frac{\sin \theta}{1-\cos \theta}=\frac{2 \sin(\theta/2)\cos(\theta/2)}{1-(1-2 \sin^2(\theta/2))}=\frac{2 \sin(\theta/2)\cos(\theta/2)}{2 \sin^2(\theta/2)}=\dots$

**maths900** · February 15th 2009, 06:09 AM

Moo, for the last bit, did you use the double angle formulas

$sin(2\theta)=2sin(\theta)cos(\theta)$

and $cos(2\theta)=1-2sin^2(\theta)$

**Soroban** · February 15th 2009, 06:17 AM

Hello, maths900!

We're expected to know the Half-Angle Identities:

. . $\sin^2\tfrac{\theta}{2} \:=\:\frac{1-\cos\theta}{2} \quad\Rightarrow\quad 1 - \cos \:=\:2\sin^2\!\tfrac{\theta}{2}$

. . $\sin\theta \:=\:2\sin\tfrac{\theta}{2} \cos\tfrac{\theta}{2}$

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

The left side is: . $\frac{\sin\theta}{1-\cos\theta} \;=\;\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{ 2} }{2\sin^2\frac{\theta}{2}} \;=\;\frac{\cos\frac{\theta}{2}}{\sin\frac{\theta} {2}}<br />$

. . . . . $= \;\frac{1}{\left(\dfrac{\sin\frac{\theta}{2}}{\cos \frac{\theta}{2}}\right)} \;=\;\frac{1}{\tan\frac{\theta}{2}}$

**maths900** · February 15th 2009, 06:31 AM

oh ok lol thanks

**Moo** · February 15th 2009, 06:34 AM

Originally Posted by maths900

Moo, for the last bit, did you use the double angle formulas

$sin(2\theta)=2sin(\theta)cos(\theta)$

and $cos(2\theta)=1-2sin^2(\theta)$

Yes :P

I prefer remembering them this way rather than "half-angle identities" :P

**maths900** · February 15th 2009, 06:43 AM

Originally Posted by Moo

Yes :P

I prefer remembering them this way rather than "half-angle identities" :P

lol me too.

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