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Math Help - trig function that passes through (x,y)

  1. #1
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    [Solved] trig function that passes through (x,y)

    hey guys! I need some help with this:

    Find the function of the form y = sin x + c that passes through the given
    points:


    1. (225, 3)
    2. (30, -2)

    i'm lost. all i know is 225 is 5pi/4 and 30 is pi/6 in radians
    Last edited by snypeshow; December 3rd 2009 at 11:22 PM.
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  2. #2
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    1. Function that passes through ( \frac{5\pi}{4},3)

    Answer: y=3\sin(x-\frac{3\pi}{4})

    Explaination:

    If you look at the normal sin graph:


    You'll notice it reaches it peak at ( \frac{\pi}{2},1)

    The amplitude is basically the height = 1/2(max-min) so here its 1/2(1-(-1)) = 1.

    You want it to reach to your y value 3, so if we look at a function, e.g. y = 3sin(x) the amplitude is always the absolute value of the # in front of the sin. That means it max is going to be 3, while its min is going to be -3.

    y=3sin(x):



    My x-scale is set to \frac{2\pi}{4}.  \frac{5\pi}{4} is half-way between  \frac{\pi}{2} (where it crosses the x-axis) and \frac{3\pi}{2}.We need to SHIFT this line to the right \frac{3\pi}{4} to get it to where we need.

    So you answer will be y=3\sin(x-\frac{3\pi}{4})


    2. You will need to put a (-) in front of sin to flip the graph. y=-2sin(x+ \frac{2\pi}{6})

    Here is the graph, divided into 6ths.
    Last edited by spoken428; December 3rd 2009 at 10:24 PM. Reason: wasn't finished
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  3. #3
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    thanks, but i don't think that's right. the equation needs to be y = sin x + c, so the c would be a vertical shift of the graph y = sin x

    if you do y = sin(x) on your graphing calc, and you set the X-Scale to 5pi/4, you'll see a sin(x) graph, and the 5pi/4 point. The graph needs to pass through the point (5pi/4, 3). the mark on the x-axis will show you where 5pi/4 is, but the y = sin(x) graph needs to have a vertical shift (upwards) to pass through that point.


    EDIT: okay, i solved the first one. the equation must pass through (5pi/4, 3) but the regular y=sin(x) graph at x = 5pi/4 is (5pi/4, -1/sqrt[2]). So if you add a vertical shift of 3 units upwards + 1/sqrt[2], the graph will pass through the point, (5pi/4, 3).
    Last edited by snypeshow; December 3rd 2009 at 10:50 PM.
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  4. #4
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    You could also use y= 3sin(x+13pi/4) as it gives you the same result.


    Graph broken down into 4ths..
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  5. #5
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    Same with the 2nd one..instead of doing the -sin you could do 2sin(x+8pi/6) and it will pass through the right point.

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  6. #6
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    no no no lol you don't understand. the equation must be in the form of y = sin x + c. that means the coefficint of sin is 1, it cannot be anything else.

    but i actually figured out the second one, logically.

    - for the graph y = sin(x): when x = pi/6, y = 0.5
    - this means there is a displacement of 0.5
    - the equation needs to pass through (pi/6, -2)
    - therefore, you need to do a vertical shift of (2 + displacement) * (-1 since its a downward shift)

    so sin(x) - 2.5 will pass through the point (pi/6, -2)

    it works because if you do sin(pi/6) you get 0.5

    I'm going to thank you anyway because you took the trouble of actually graphing it on a ticalc emulator and posted it up. thanks for your help!!!
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  7. #7
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    crazyness..haha at least you got it!
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