• Jun 18th 2009, 01:13 PM
sophiaroth
secx = (2(cosxsin2x - sinxcos2x))/sin2x
• Jun 18th 2009, 01:26 PM
skeeter
Quote:

Originally Posted by sophiaroth
secx = (2(cosxsin2x - sinxcos2x))/sin2x

$\displaystyle \frac{2[\cos{x}\sin(2x) - \sin{x}\cos(2x)]}{\sin(2x)}$

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

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

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

$\displaystyle \frac{1}{\cos{x}} = \sec{x}$
• Jun 18th 2009, 01:36 PM
sophiaroth
thanks! i had tried cos2x = cos^2x - sin^2x, and never went as far as 2cos^2x - 1
• Jun 18th 2009, 10:55 PM
great_math
Quote:

Originally Posted by sophiaroth
secx = (2(cosxsin2x - sinxcos2x))/sin2x

$\displaystyle \text{this is sin(2x-x)}$
$\displaystyle \frac{2\overbrace{(\cos{x}\sin(2x) - \sin{x}\cos(2x))}}{\sin(2x)}$

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

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

just to get rid of few steps :)