User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)

Thoughts on the picture, anyone? --GFauxPas 14:37, 9 November 2011 (CST)


 * Looks okay to me. I was going to get round to doing something similar myself at one point.


 * Mind, if you're planning on using it to illustrate sine and cosine, you might want to add the actual distances as lines. Oh, and purists may wince when they see x and y used to define the axes and the point on it, but I wouldn't be too fussed. --prime mover 14:47, 9 November 2011 (CST)

Try 2. I see I lost too many colors by saving it as a .gif, try 3 will be a .png or something.

File:Unitcirclev2.gif

\Let $P = (x,y)$ be a point on the unit circle centered at the origin.

Let $\theta$ be the angle formed by the arc $(1,0)$, $(x,y)$ subtending the origin, measured counterclockwise.

The unit circle definition of the trigonometric functions are $\cos \theta := x$

$\sin \theta := y$

That is, the directed distance between $P$ and the $x$-axis is the cosine, and the directed distance between $P$ and the $y$-axis is the sine.

Sources: khan academy "tau versus pi", wolfram mathworld "trigonometry"

After this is set up I can do a proof of the consistency between the right triangle definition and the circle definition. --GFauxPas 07:22, 17 November 2011 (CST)

What to Call this Theorem?
Hello friends, I'd like to add a proof for:


 * $\displaystyle \int \frac {1}{x^2 + a^2} \ \mathrm dx$

but I don't know what to call it. "Integral of a Rational Function"? "Integral Involving Arctangent"? "Integral of 1 Over (x^2 + a^2)"? What should I name the page? --GFauxPas 06:55, 15 December 2011 (CST)

Moved to Integral Involving Arctangent, please change the name if you think of something better. --GFauxPas 14:02, 15 December 2011 (CST)

Theorem

 * $\displaystyle \int \frac {\mathrm dx}{\sqrt{a^2 - x^2}} \ \mathrm dx = \arcsin{\frac{x}{a}} + C$

where $a$ is a constant and $a^2 > x^2$.

Proof
Substitute:


 * $\sin \theta = \dfrac {x}{a} \iff x = a \sin \theta$

for $\theta \in \left[-\dfrac {\pi}{2}..\dfrac {\pi}{2}\right]$.

From Shape of Sine Function, this substitution is valid for all $\dfrac{x}{a} \in \left[-1..1\right]$.

By hypothesis:


 * $a^2 > x^2$

$\implies 1 > \left( \dfrac {x}{a} \right)^2$

$\implies 1 > \left|\dfrac{x}{a}\right|$

$\implies -1 < \dfrac {x}{a} < 1$

... so this substitution will not change the domain of the integrand.

I'll finish this tomorrow, it's late. Also, I can probably make those inequalities look better, but I'm done for the night. --GFauxPas 23:03, 15 December 2011 (CST)