Finding exact values for trigonometry

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May 13th 2008, 03:36 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, 05:12 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

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