1. Trig Identity

Hey everyone, i'm having some troubles figuring out this trig identity!

DaCoo911

2. Originally Posted by DaCoo911
Hey everyone, i'm having some troubles figuring out this trig identity!

DaCoo911

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

3. Hello,
Originally Posted by DaCoo911
Hey everyone, i'm having some troubles figuring out this trig identity!

DaCoo911

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

for the denominator, use the following identities :
$(a-b)(a+b)=a^2-b^2$
$\cos^2(x)+\sin^2(x)=1$

4. Originally Posted by DaCoo911
Hey everyone, i'm having some troubles figuring out this trig identity!

DaCoo911

Multiply the LHS by $\frac{1-cos(x)}{1-cos(x)}$ - this will give is the difference of two squares on the bottom:

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

cancelling a sin(x):

$\frac{(1-cos^2(x))}{sin(x)}$ which equals the RHS

5. Originally Posted by DaCoo911
Hey everyone, i'm having some troubles figuring out this trig identity!

DaCoo911

Hi DaCoo911,

Here's another approach. Work with the left hand side.

Multiply the left side by $\frac{\sin x}{\sin x}=$

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

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

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

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