cosx = -5/13 ... find sin2x

**mike41** · December 10th 2008, 04:54 PM

cosx = -5/13 where pi/2 < x < pi find

a) sin2x

any idea how to do this?

**skeeter** · December 10th 2008, 05:06 PM

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 ...

$2\sin{x} \cos{x} = \sin(2x)$

**mike41** · December 10th 2008, 05:24 PM

wow thank you skeet

thought there was another way to solve it

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