1. ## Trig Identities

Hi,
I have my math exam in a few days and that's the last question I have to solve to complete the trig part.
$2sin x(1+cos x) = \frac {2sin^3 x} {1-cos x}$
I've been trying to use sin^2x + cos^2x for LS and RS however I'm getting nowhere with that, mainly because of the right side fraction.
Any help appreciated.
Thank you very much!

2. $2\sin x(1 + \cos x) = 2\sin x(1 + \cos x) \cdot \frac{{1 - \cos x}}
{{1 - \cos x}} = \frac{{2\sin x\left( {1 - \cos ^2 x} \right)}}
{{1 - \cos x}}.$

The rest follows.

3. I see...
So you are allowed to multiply 1-cos x over 1-cos x at any time?
I mean if I'd get anything similar in my exam I can just multiply it like that?
Thanks

4. Originally Posted by Universe
I see...
So you are allowed to multiply 1-cos x over 1-cos x at any time?
It's like multiplyin' by 1. It's like $\frac aa.$

Originally Posted by Universe
I mean if I'd get anything similar in my exam I can just multiply it like that?
That depends of the problem.

For example your trig. identity could've been tackled by workin' the RHS.

5. Work with the right side and show it equals the left side.

Multiply top and bottom of right side by $1+cos(x)$

$\frac{2sin^{3}(x)}{1-cos(x)}\cdot\frac{1+cos(x)}{1+cos(x)}$

$=\frac{2sin^{3}(x)+2sin^{3}(x)cos(x)}{sin^{2}(x)}$

$2sin(x)+2sin(x)cos(x)$

Factor:

$2sin(x)(1+cos(x))$

6. Alright.
Thank you very much guys.