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Thread: Finding exact values for trigonometry

  1. May 13th 2008, 03:36 PM #1
    kelsey3
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    Finding exact values for trigonometry

    Find the exact value of sin2θ given that cosθ= -12/13 and θ is in Quadrant II.
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  2. May 14th 2008, 05:12 AM #2
    topsquark
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    Quote Originally Posted by kelsey3 View Post
    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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