Math Help - Trig Proof, help please!

1. Trig Proof, help please!

secx = (2(cosxsin2x - sinxcos2x))/sin2x

2. Originally Posted by sophiaroth
secx = (2(cosxsin2x - sinxcos2x))/sin2x
$\frac{2[\cos{x}\sin(2x) - \sin{x}\cos(2x)]}{\sin(2x)}$

$\frac{2[\cos{x} \cdot 2\sin{x}\cos{x} - \sin{x}(2\cos^2{x}-1)]}{2\sin{x}\cos{x}}$

$\frac{2[2\cos^2{x}\sin{x} - 2\cos^2{x}\sin{x} + \sin{x}]}{2\sin{x}\cos{x}}$

$\frac{2\sin{x}}{2\sin{x}\cos{x}}$

$\frac{1}{\cos{x}} = \sec{x}$

3. thanks! i had tried cos2x = cos^2x - sin^2x, and never went as far as 2cos^2x - 1

4. Originally Posted by sophiaroth
secx = (2(cosxsin2x - sinxcos2x))/sin2x
$\text{this is sin(2x-x)}$
$\frac{2\overbrace{(\cos{x}\sin(2x) - \sin{x}\cos(2x))}}{\sin(2x)}$

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

$=\frac{1}{\cos{x}} = \sec{x}$

just to get rid of few steps