1. ## Verifying identities

I'm a little confused how to start this problem:

1 + cosx + sinx
sinx 1 + cosx

2. Originally Posted by Lauren Schmidt
I'm a little confused how to start this problem:

1 + cosx + sinx
sinx 1 + cosx
Lauren you have to be a little more clear on your problem. Are we proving this identity to be something else? It just looks like a whole lot of cos's and sin's thrown onto the screen at the moment.

Could it be like this?

$\frac{1+cos(x)+sin(x)}{sin(1+cos(x))} = \cdots$

3. oh goodness. haha. i'm sorry. it's = to 2 cscx

1 + cosx + sinx = 2 cscx
sinx 1 + cosx

4. Originally Posted by Lauren Schmidt
oh goodness. haha. i'm sorry. it's = to 2 cscx

1 + cosx + sinx = 2 cscx
sinx 1 + cosx
So maybe?

$\frac{1+cos(x)+sin(x)}{sin(1+cos(x))} = 2 cosec(x)$

5. Looks like a tricky one, these proofs can be unforgiving at times but if you get one to fall out it feels great.

Have a look at this list of identities

Table of Trigonometric Identities

Then remember that

$\frac{1}{sin(x)} = cosec(x)$

Try try to put everything one the left hand side in terms of sin(x)

6. Originally Posted by Lauren Schmidt
oh goodness. haha. i'm sorry. it's = to 2 cscx

1 + cosx + sinx = 2 cscx
sinx 1 + cosx

(1+cosx)/sinx + sinx/(1+cosx)

$\frac{1+\cos{x}}{\sin{x}} + \frac{\sin{x}}{1+\cos{x}} = 2\csc{x}$

$\frac{1+\cos{x}}{\sin{x}} + \frac{\sin{x}}{1+\cos{x}} \cdot \frac{1-\cos{x}}{1-\cos{x}} =$

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

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

$\frac{1+\cos{x}}{\sin{x}} + \frac{1-\cos{x}}{\sin{x}} =$

$\frac{1+\cos{x} + 1 - \cos{x}}{\sin{x}} =$

$\frac{2}{\sin{x}} = 2\csc{x}$