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Intuition why area of a period of sinx =4 = area of square unit circle

  1. Jul 7, 2013 #1
    1. The problem statement, all variables and given/known data

    This isn't really homework, but I've been reviewing calc & trig and realized that the area of one period of sin(x) = 4. Since sin(θ) can be understood as the y-value of points along a unit circle, I noticed that the area of a unit square that bounds the unit circle is also 4. Is this a relationship about squaring a circle, or just a coincidence?

    2. Relevant equations

    A unit circle is a circle with a radius of one.

    Area of one period of sin x is [itex] 2 \int^{\pi}_{0} sin(x) dx = 4[/itex]

    For a unit circle, r=1. So the area of a square bounding the unit circle is also [itex](2r)^{2} = 4[/itex].

    Trigonometry_700.gif Sin_550.gif

    3. The attempt at a solution

    I tried drawing out what the area under the curve of sin(x) means, focusing on the first quarter of the unit circle (so, from 0 to pi/2, which is 1/4 the period of sinx and has an area of 1. The square bounding the quarter of a circle also has an area of [itex]r^{2}=1[/itex].)

    I understand that the area under sin(x) is the infinite sum of all measurements of the y-coordinate of a point on a rotating unit circle. But why does that become a square?

    In other words, what does the area under sin(x) mean and what is its relationship to the square bounding the unit circle (or the 1x1 square bounding the quarter of the circle)? Why?

    I hope I have conveyed this question clearly. Thank you for your help!
     
    Last edited: Jul 7, 2013
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  3. Jul 7, 2013 #2

    Simon Bridge

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    Niggle: The area between the x axis and sin(x) for any integer number of periods is 0.
    What you did was the area of |sin(x)| ...

    The relationship is to do with the way "sine" is defined on the unit circle.
    You can think of it like the way Pythagoras sometimes gets demonstrated by putting squares on each side of a right-angle triangle and showing that the two smaller squares can be cut up so they fit exactly inside the biggest one.

    Note: does it make a difference if the circle has unit circumference instead of unit area?
     
  4. Jul 7, 2013 #3
    I don't get what you mean about the cut up triangles for this application. My confusion is that it doesn't fit. The area under a period of |sin x| = 4, a unit circle's area is only [itex]\pi[/itex].

    A unit circle is a circle with a radius of one, not an area of one. I've now made this explicit in the question.
     
  5. Jul 7, 2013 #4

    Simon Bridge

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    Its a simile - an analogy ...
    The area under the sine from 1 to pi/2 can be cut up to fill the gap between the sin and y=1 from 0 to 1.

    That was already clear. The sine is a specific length defined on the unit-radius circle... since the one was derived from the other, it is not surprising to find they have special relationships.
     
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