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Math Help - Proofs for common formulas in trig and calculus

  1. #1
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    Proofs for common formulas in trig and calculus

    It frustrates me to be presented with a formula in a math book, but there is no proof to show that it is true.
    Does anyone have proofs for the following?
    The length of an arc of a circle, s, subtended by a central angle, theta, and a radius, r is: theta = s/r -- how to prove this?
    How to prove that the circumference of a circle = 2*pi*r
    How to prove area of circle = pi*r^2
    Thanks
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  2. #2
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    Re: Proofs for common formulas in trig and calculus

    Quote Originally Posted by estad View Post
    It frustrates me to be presented with a formula in a math book, but there is no proof to show that it is true.
    Does anyone have proofs for the following?
    The length of an arc of a circle, s, subtended by a central angle, theta, and a radius, r is: theta = s/r -- how to prove this?
    How to prove that the circumference of a circle = 2*pi*r
    How to prove area of circle = pi*r^2
    Thanks
    All three can be done by realizing that all circles are the same shape so "similar". A little more technical way of saying that is that since circumference is a distance measurement, just as diameter is, circumference is proportional to diameter- circumference is a multiple of diameter- and we define \pi to be that multiple. One way to prove the area formula, that goes back to Archimedes, is to divide the circle into n wedges with equal interior angle, then "alternate" them so that the first has its "point" downward, the next upward, the next downward and so on. That gives a figure that "looks like" a rectangle with a "wavy" edge. That rectangle would have one side of length r and the other half the circumference, \pi r, and so area \pi r^2. Take the limit as n goes to infinity and the thing that "looks like" a rectangle becomes a rectangle.

    Of course, a "modern" way to do this would be to use an arclength integral for the circumference and an area integral for the area.
    If you let x= r cos(t), y= r sin(t), with t from 0 to 2\pi then x^2+ y^2= r^2 so those are parametric equations for the circle of radius r. The circumference is given by \int_0^{2\pi}\sqrt{r^2cos^2(t)+ r^2sin^2(t)}dt= \int_0^{2\pi}r dt= 2\pi r and the area is given by \int_{-r}^r 2\sqrt{r^2- x^2}dx. Let x=r sin(t) so that dx=r cos(t)dt and that becomes 2 \int_{-\pi/2}^{\pi/2} r^2\sqrt{1- sin^2(t)}(cos(t)dt)= 2r^2 \int_{-\pi/2}^{\pi/2} cos^2(t)dt. cos^2(t)= \frac{1}{2}(1+ cos(2t)) so we can write that as r^2 \int_{-\pi/2}^{\pi/2} (1+ cos(2t))dt= \left[r^2 (t+ \frac{1}{2}sin(2t)\right]_{-\pi/2}^{\pi/2}

    The "sin(2t)" evaluated at \pi/2 and -\pi/2 is 0. That leaves r^2(\pi/2- (-\pi/2)= \pi r^2.

    One might argue that invalid to use sine and cosine in parametric equations or to use the sine substitution for the area integral because those functions are themselves defined in terms of distance around a circle but that is not true. It is perfectly valid to define "sin(x)" as the power series \sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!} and define "cos(x)" as the power series \sum_{n=0}^\infty \frac{x^{2n}}{(2n)!}. Or, equivalently, define sin(x) to be the solution to the differential equation y''= -y with initial values y(0)= 0, y'(0)= 1 and define cos(x) to be the solution to the differential equation y''= -y with initial values y(0)= 1, y'(0)= 0.
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    Re: Proofs for common formulas in trig and calculus

    Thanks for your answer, HallsofIvy.
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    Re: Proofs for common formulas in trig and calculus

    To answer your first question - angles represent a proportion of a circle swept out. Either in degrees as "how many 360ths of the circle have been swept out" or in radians as "what proportion of the $\displaystyle \begin{align*} 2\pi \end{align*}$ lengths of the radius have been swept out along the circumference?"

    Radians are easier to use in the arclength formula (though not essential). If you remember that the angle $\displaystyle \begin{align*} \theta \end{align*}$ represents the proportion of $\displaystyle \begin{align*} 2\pi \end{align*}$ has been swept out along the circumference, then

    $\displaystyle \begin{align*} l &= \frac{\theta}{2\pi} \cdot C \\ &= \frac{\theta}{2\pi} \cdot 2\pi\, r \\ &= \theta \, r \end{align*}$
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