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Math Help - A trig question

  1. #1
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    A trig question

    Hi guys

    Just have a question about the sin function

    If the function is given as sin2(x).

    How would i know where it will cut the x-axis (assuming the x axis is denoted as using 'pi')

    Cause in the original function: sin(x) if we put in 'pi' into x then it will equal 0.

    So now, is there an approach that given the function: sin2(x), that i could know where it cuts the x-axis ie when y = 0?

    thanks in advance guys


    Forgot to mention:

    How can i also find the trig. function values.

    Like

    sin when (-pi)/4 the value is -squareroot x divded by 2.

    How can i also use that to figure out a value like pi/4 given the function sin2x
    Last edited by Redeemer_Pie; June 9th 2009 at 03:26 AM. Reason: forgot to ask a question
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  2. #2
    Super Member
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    I'm assuming you mean y = \sin 2x and not y = {\sin}^2 x.

    If you know the graph of the basic sine function, you would know that the x-intercepts are x = n\pi, for any integer n (in other words, multiples of pi). Putting a number greater than 1 in front of x transforms the graph -- you have a horizontal shrink by a factor of that number. So in our case, y = \sin 2x is a horizontal shrink of y = {\sin}^2 x by a factor of 2.

    This effectively means that the x-intercepts occur twice as often, ie. x = \frac{n\pi}{2}, for any integer n.

    Another example: say it's y = \sin 4x. Then its x-intercepts would be x = \frac{n\pi}{4}, for any integer n.


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  3. #3
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    Thanks a lot!!!!

    Yes sin2x is not sin to the power of 2 x

    So if i graphed the function sin2x, the first point it will cut the x-axis would be 'pi'/2?

    As well if it was like sin4x then the first point it will cut will be 'pi'/4?

    Thanks again mate
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