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Math Help - Circular Functions

  1. #1
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    Circular Functions

    Hey Ive got two questions regarding Cicular Functions, could you possibly include some working as I am really struggling with this topic:

    I need to find the next four values for (t):
    6=3sin(2t)+3
    I know the first one is 5(pie)/4, i think the second is 9(pie)/4

    and
    between (pie) and 4(pie)
    y=2sin(3t-(pie)/4)+8 for which y attains the minimum value.

    thanks guys
    RoboStar
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  2. #2
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    Quote Originally Posted by RoboStar View Post
    Hey Ive got two questions regarding Cicular Functions, could you possibly include some working as I am really struggling with this topic:

    I need to find the next four values for (t):
    6=3sin(2t)+3
    I know the first one is 5(pie)/4, i think the second is 9(pie)/4
    Yes I think you are right

    between (pie) and 4(pie)
    y=2sin(3t-(pie)/4)+8 for which y attains the minimum value.

    thanks guys
    RoboStar
    y=2sin(3t-(pie)/4)+8

    Minimum value of \sin function is -1. So minimum value of y could be 6.The real question is does there exist a t between (pie) and (4pie), such that this happens.
    sin(3t-(pie)/4) = -1 means that 3t - \frac{\pi}{4} = -\frac{\pi}2,\frac{3\pi}2,\frac{7\pi}2,....

    Thus 3t = -\frac{\pi}4,\frac{7\pi}2,\frac{15\pi}2,....

    So t = \frac{7\pi}6,\frac{15\pi}6,....

    In general t = \frac{(8n-1)\pi}6, n \in \mathbb{N}

    So between \pi an 4\pi, we have three angles that satisfy:

    t = \frac{7\pi}6,\frac{15\pi}6,\frac{23\pi}6 and the minimum value is y = 6<br /> <br />
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  3. #3
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    and for the first question, would the next two be 13(pie)/4 and 17(pie)/4? I'm still quite unsure with this concept.
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  4. #4
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    Quote Originally Posted by RoboStar View Post
    [snip]
    I need to find the next four values for (t):
    6=3sin(2t)+3
    I know the first one is 5(pie)/4, i think the second is 9(pie)/4

    [snip]
    Do you need values such that t > 0. If so, you missed t = pi/4

    Note that your equation can be re-arranged into sin(2t) = 1.

    The general solution is 2t = \frac{\pi}{2} + 2n \pi \Rightarrow t = \frac{\pi}{4} + n \pi where n is an integer.

    So get all the solutions you want by letting n = 0, 1, 2, .... -1, -2, ....

    Quote Originally Posted by RoboStar View Post
    and for the first question, would the next two be 13(pie)/4 and 17(pie)/4? I'm still quite unsure with this concept.
    Yes (they correspond to n = 3 and n = 4).
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