# Finding exact values for trigonometry

• May 13th 2008, 04:36 PM
kelsey3
Finding exact values for trigonometry
Find the exact value of sin2θ given that cosθ= -12/13 and θ is in Quadrant II.
• May 14th 2008, 06:12 AM
topsquark
Quote:

Originally Posted by kelsey3
Find the exact value of sin2θ given that cosθ= -12/13 and θ is in Quadrant II.

$cos(\theta) = -\frac{12}{13}$
tells us that $sin(\theta) = \pm \sqrt{1 - cos^2(\theta)} = \pm \frac{5}{13}$

Since we know that $\theta$ is in QII where sine is positive, we know to use the + sign.

Now
$sin(2\theta) = 2~sin(\theta)~cos(\theta)$
and you can take it from there.

-Dan