# Prove this

• Sep 14th 2009, 06:17 PM
jasonlewiz
Prove this
Prove this guys,.,thanks

2 csc θ = 1 - cos θ + sin θ
--------- --------
sin θ 1-cos θ
• Sep 14th 2009, 08:56 PM
seld
is that supposed to be :

$\displaystyle \frac{2csc\theta}{sin\theta}=\frac{1-cos\theta+sin\theta}{1-cos\theta}$ ?
• Sep 14th 2009, 10:25 PM
CaptainBlack
Quote:

Originally Posted by seld
is that supposed to be :

$\displaystyle \frac{2csc\theta}{sin\theta}=\frac{1-cos\theta+sin\theta}{1-cos\theta}$ ?

But that is not an identity.

Looking at the source it appears that:

$\displaystyle 2\csc (\theta)= 1 -\frac{\cos(\theta)}{\sin(\theta)} +\frac{\sin(\theta)}{1-\cos(\theta)}$

is the indended expression, but that is also not an identity.

CB
• Sep 14th 2009, 10:29 PM
seld
Quote:

Originally Posted by CaptainBlack
But that is not an identity.

CB

true, but that's all I could make of the original post with the

----- -------
sin x 1-cosx

doesn't make any sense to me to have (sin x)*1-cos x as a denominator.

looking at the source? err . . . the 2 dissapeared . . .
• Sep 15th 2009, 01:36 AM
CaptainBlack
Quote:

Originally Posted by seld
true, but that's all I could make of the original post with the

----- -------
sin x 1-cosx

doesn't make any sense to me to have (sin x)*1-cos x as a denominator.

looking at the source? err . . . the 2 dissapeared . . .

If you had quoted what the OP had posted you will have seen what they typed, and been better able to interpret it.

I have put the lost 2 back where it belongs now.

CB