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Thread: Finding values of t for sin&tan

  1. #1
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    Finding values of t for sin&tan

    I have homework where i need to do the following:

    3. For what values of t on the interval [0,2π] is: a. sin(t) = -1/2
    b. tan(t) = -
    √3

    *in part 'b' the symbol is a square root, meaning the negative square root of 3*


    I think it can be solved using the unit circle (..?) but I can't figure it out.
    Help?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Dust View Post
    I have homework where i need to do the following:

    3. For what values of t on the interval [0,2π] is: a. sin(t) = -1/2
    b. tan(t) = -
    √3

    *in part 'b' the symbol is a square root, meaning the negative square root of 3*


    I think it can be solved using the unit circle (..?) but I can't figure it out.
    Help?
    you should know by heart that \sin \frac {\pi}6 = \frac 12 and \tan \frac {\pi}3 = \sqrt{3}

    now, use your knowledge of reference angles to continue

    can you?
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  3. #3
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    I know that sin pi/6 is 1/2 and tan pi/3 is the square root of 3.
    I understand what you said.

    My problem is the negative signs are throwing me off.
    Would the sine graph be 11π/6 and 7π/6 ??
    And the tan graph would be 2π/3 and 5π/3 ??

    If not i am truly lost.
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  4. #4
    Super Member 11rdc11's Avatar
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    sin is negative in the 3rd and 4th quad and tan is negative in the 2nd and 4th quad
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  5. #5
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    So i am correct?

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  6. #6
    A riddle wrapped in an enigma
    masters's Avatar
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    Quote Originally Posted by Dust View Post
    So i am correct?

    Yep!!

    But what about \tan(t)=-\sqrt{3}?

    The reference angle in the 2nd quadrant where tan is negative is .

    The reference angle in the 4th quadrant where tan is negative is .
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