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

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- May 13th 2008, 03:36 PMkelsey3Finding exact values for trigonometry
Find the exact value of sin2θ given that cosθ= -12/13 and θ is in Quadrant II.

- May 14th 2008, 05:12 AMtopsquark
$\displaystyle cos(\theta) = -\frac{12}{13}$

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

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

Now

$\displaystyle sin(2\theta) = 2~sin(\theta)~cos(\theta)$

and you can take it from there.

-Dan