1. ## identity problems

Couple of problems from todays homework that i'm struggling with. Thanks for having a look.

$\displaystyle \frac{sinx+tanx} {1+cos}=\frac{sin2x}{2cos^2x}$

$\displaystyle \frac{sin2x}{2-2cos^2x}=cotx$

2. Originally Posted by nightrider456
Couple of problems from todays homework that i'm struggling with. Thanks for having a look.

$\displaystyle \frac{sinx+tanx} {1+cos}=\frac{sin2x}{2cos^2x}$

$\displaystyle \frac{sin2x}{2-2cos^2x}=cotx$
$\displaystyle \frac{\sin{x} + \tan{x}}{1 + \cos{x}} = \frac{\sin{x} + \frac{\sin{x}}{\cos{x}}}{1 + \cos{x}}$

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

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

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

$\displaystyle = \frac{\sin{x}}{\cos{x}}$

$\displaystyle = \frac{\sin{x}\cos{x}}{\cos^2{x}}$

$\displaystyle = \frac{\frac{1}{2}\sin{2x}}{\cos^2{x}}$

$\displaystyle = \frac{\sin{2x}}{2\cos^2{x}}$.

3. Originally Posted by nightrider456
Couple of problems from todays homework that i'm struggling with. Thanks for having a look.

$\displaystyle \frac{sinx+tanx} {1+cos}=\frac{sin2x}{2cos^2x}$

$\displaystyle \frac{sin2x}{2-2cos^2x}=cotx$
$\displaystyle \frac{\sin{2x}}{2 - 2\cos^2{x}} = \frac{2\sin{x}\cos{x}}{2 - 2\cos^2{x}}$

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

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

$\displaystyle = \frac{\sin{x}\cos{x}}{\sin^2{x}}$

$\displaystyle = \frac{\cos{x}}{\sin{x}}$

$\displaystyle = \cot{x}$