# Proving trig identities

• Mar 31st 2013, 10:37 PM
camjerlams
Proving trig identities
Hi, I've been working on a question for ages and can't find an answer. This is the first proof I've been asked to do also.

prove:

$\displaystyle \frac{\sin(x)}{1-\cos(x)} = \csc(x) + \cot(x)$

Am I right in to work the LHS like this:

$\displaystyle \sin(x) = \sqrt{1-\cos^2(x)}$

So,

$\displaystyle \sqrt{1-\cos^2(x)} = 1-\cos(x)$ ..? For $\displaystyle \frac{\sin(x)}{\sin(x)}$

So LHS =1?
And what do do with the RHS?
• Apr 1st 2013, 12:24 AM
princeps
Re: Proving trig identities
Hint :

Multiply both numerator and denominator of the LHS fraction by :

$\displaystyle 1+\cos x$
• Apr 1st 2013, 04:54 AM
camjerlams
Re: Proving trig identities
Bigger hint?

So,

$\displaystyle \frac{\sin(\Theta)}{1-cos(\Theta)}\times\frac{1+cos(\Theta)}{1+cos(\Thet a)}=\frac{sin(\Theta)\times(1+\cos(\Theta)}{1-\cos^2(\Theta)}=\frac{1+\cos(\Theta)}{1-\cos(\Theta)}$...
• Apr 1st 2013, 04:59 AM
Prove It
Re: Proving trig identities
Quote:

Originally Posted by camjerlams
Bigger hint?

So,

$\displaystyle \frac{\sin(\Theta)}{1-cos(\Theta)}\times\frac{1+cos(\Theta)}{1+cos(\Thet a)}=\frac{sin(\Theta)\times(1+\cos(\Theta)}{1-\cos^2(\Theta)}=\frac{1+\cos(\Theta)}{1-\cos(\Theta)}$...

Actually

\displaystyle \displaystyle \begin{align*} \frac{\sin{(\theta)} \left[ 1 + \cos{(\theta)} \right] }{1 - \cos^2{(\theta)}} &\equiv \frac{\sin{(\theta)} \left[ 1 + \cos{(\theta)} \right] }{ \sin^2{(\theta)} } \\ &\equiv \frac{1 + \cos{(\theta)}}{\sin{(\theta)}} \\ &\equiv \frac{1}{\sin{(\theta)}} + \frac{\cos{(\theta)}}{\sin{(\theta)}} \\ &\equiv \csc{(\theta)} + \cot{(\theta)} \end{align*}
• Apr 1st 2013, 05:39 AM
camjerlams
Re: Proving trig identities
Ah yeah I totally didn't see the sin^2(x), thanks for that(Nod)