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Math Help - Determining the rule for graphs of trig function

  1. #1
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    Determining the rule for graphs of trig function

    A few question i am stuck on:

    1) y=Atan(nt) find A and n using the information provided
    Asymptotes have equation t=(2k+1)\frac{\pi}{6}where k=>Z
    when t=\frac{\pi}{12},y=5

    2) y=Asin(nt+e)
    Range = [-2,2]
    Period=6
    When t=1,y=1

    I have found what A and n is however i don't know how to find e.

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    A few question i am stuck on:

    1) y=Atan(nt) find A and n using the information provided
    Asymptotes have equation t=(2k+1)\frac{\pi}{6}where k=>Z
    when t=\frac{\pi}{12},y=5

    2) y=Asin(nt+e)
    Range = [-2,2]
    Period=6
    When t=1,y=1

    I have found what A and n is however i don't know how to find e.

    P.S
    2. You should know that for a sine function of the form

    y = a\sin{(bx)},

    the amplitude is a and the period is \frac{2\pi}{b}.


    y = A\sin{(nt + e)}.

    Since the range is [-2, 2], that means the amplitude must be 2.

    Therefore A = 2.


    The period is 6.

    So 6 = \frac{2\pi}{n}

    n = \frac{2\pi}{6}

    n = \frac{\pi}{3}.


    So far you have

    y = 2\sin{\left(\frac{\pi t}{3} + e\right)}.


    You also know that when t = 1, y= 1.

    So 1 = 2\sin{\left(\frac{\pi}{3} + e\right)}

    \frac{1}{2} = \sin{\left(\frac{\pi}{3} + e\right)}

    \frac{\pi}{6} = \frac{\pi}{3} + e

    -\frac{\pi}{6} = e.


    Therefore y = 2\sin{\left(\frac{\pi t}{3} - \frac{\pi}{6}\right)}.
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