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

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

    2)$\displaystyle y=Asin(nt+e)$
    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

    $\displaystyle y = a\sin{(bx)}$,

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


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

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

    Therefore $\displaystyle A = 2$.


    The period is 6.

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

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

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


    So far you have

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


    You also know that when $\displaystyle t = 1, y= 1$.

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

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

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

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


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