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    Laken 7531's Avatar
    Laken 7531 Posts: 3, Reputation: 1
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    #1

    Sep 24, 2008, 03:47 PM
    graping sin/cos
    How to graph: y=1-3sin2x
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #2

    Sep 25, 2008, 01:50 AM

    Do you know how to graph y=sin(x) first? If you do, then we'll continue. I must be sure you know that first, for the problem to be solved.
    Laken 7531's Avatar
    Laken 7531 Posts: 3, Reputation: 1
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    #3

    Sep 25, 2008, 06:04 AM

    yes I do know how to graph y=sin(x).
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #4

    Sep 26, 2008, 02:19 AM

    Ok, then, let's go to sin(2x). This '2' in the brackets in front of the 'x' means that there are two cycles of the graph in 360 degrees.

    In other words, it's as if you 'compressed' the graph to make it shorter. In 360 degrees, then you'll have 2 max points and 2 min points instead of only one in the usual sin(x) graph.

    Then, we come to the '3' in front of the sin(2x). That means that you need to make your graph longer, along the vertical axis. In sin(2x), your first max point is at (45, 1). In the graph of 3sin(2x), the max point rises up to three times the y-coordinate, that is at (45, 3). The min point, originally at (90, -1) is found to be at (90, -3). The same applies to the next max and min points.

    That is already a big part. If you are OK with that, we'll continue with the -ve sign in front of the '3sin(2x)' then to the '1'. I must be sure you're OK with what I'm trying to tell you.

    By the way, a side exercise;
    Give the coordinates of the second maximum point and the second miminum point.
    Laken 7531's Avatar
    Laken 7531 Posts: 3, Reputation: 1
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    #5

    Sep 26, 2008, 05:58 AM

    well, all I needed to know was how to graph the 3sin2x part I already know how to graph the negative sign and the 1 in front I was just confused on how to do that. Thank you, you helped a lot
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #6

    Sep 27, 2008, 02:27 AM

    You're most welcomed and I'm glad to have been able to help you. I hope to see you again:)!

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