Hey everyone, i'm having some troubles figuring out this trig identity!
Thanks in advanced!
DaCoo911
Hello,
$\displaystyle \frac{\sin(x)}{1+\cos(x)}=\frac{\sin(x)}{1+\cos(x) } \cdot \frac{1-\cos(x)}{1-\cos(x)}=\frac{\sin(x)(1-\cos(x))}{(1+\cos(x))(1-\cos(x))}$
for the denominator, use the following identities :
$\displaystyle (a-b)(a+b)=a^2-b^2$
$\displaystyle \cos^2(x)+\sin^2(x)=1$
Multiply the LHS by $\displaystyle \frac{1-cos(x)}{1-cos(x)}$ - this will give is the difference of two squares on the bottom:
$\displaystyle \frac{sin(x)(1-cos(x))}{1-cos^2(x)} = \frac{sin(x)(1-cos^2(x))}{sin^2(x)}$
cancelling a sin(x):
$\displaystyle \frac{(1-cos^2(x))}{sin(x)}$ which equals the RHS
Hi DaCoo911,
Here's another approach. Work with the left hand side.
Multiply the left side by $\displaystyle \frac{\sin x}{\sin x}=$
$\displaystyle \frac{\sin^2 x}{\sin x(1+\cos x)}=$
$\displaystyle \frac{1-\cos^2 x}{\sin x(1+ \cos x)}=$
$\displaystyle \frac{(1-\cos x)(1+ \cos x)}{\sin x(1+ \cos x)}=$
$\displaystyle \frac{1-\cos x}{\sin x}$