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Math Help - trigonometric function values

  1. #1
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    Unhappy trigonometric function values

    find sin (theta).

    tan(theta)= - (sqrt)7/2

    sec(theta)>0

    I'm sorry if this is confusing to you all(math editor) I am also very confused! Please help!
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  2. #2
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    You can solve for theta by taking the arctan(-\frac{\sqrt(7)}{2}). Then just plug in theta as the parameter for sin().
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  3. #3
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    Quote Originally Posted by momuvfour View Post
    find sin (theta).

    tan(theta)= - (sqrt)7/2

    sec(theta)>0

    I'm sorry if this is confusing to you all(math editor) I am also very confused! Please help!
    Dear momuvfour,

    It is given that sec\theta>0\Rightarrow{cos\theta<0}That is theta is in the second or third quadrants. Hence we have to find values for tan\theta=-\frac{\sqrt{7}}{2} such that theta will be in the second or third quadrants. Can you continue from here??

    Dear macosxnerd101,

    By taking \theta=arctan\left(-\frac{\sqrt{7}}{2}\right) you have only considered the case where theta is in the second quadrant.

    Hope this will help you.
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