# cosx = -5/13 ... find sin2x

• December 10th 2008, 05:54 PM
mike41
cosx = -5/13 ... find sin2x
cosx = -5/13 where pi/2 < x < pi find

a) sin2x

any idea how to do this?
• December 10th 2008, 06:06 PM
skeeter
find the value of $\sin{x}$ using the value they gave you for $\cos{x}$ by sketching a reference triangle in quad II.

then use the double angle identity ...

$2\sin{x} \cos{x} = \sin(2x)$
• December 10th 2008, 06:24 PM
mike41
wow thank you skeet

thought there was another way to solve it