I am looking at the unit circle proof, and I cannot understand how abs(sinx-sinxo) is less than or equal to abs(x-xo). How is abs(x-xo) considered the length of the arc portion?

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- May 18th 2010, 11:33 AMseams192Can anyone prove sin(x) continuous
I am looking at the unit circle proof, and I cannot understand how abs(sinx-sinxo) is less than or equal to abs(x-xo). How is abs(x-xo) considered the length of the arc portion?

- May 18th 2010, 11:40 AMPlato
- May 19th 2010, 05:09 AMHallsofIvy
But this thread was titled "Can anyone prove sin(x) continuous". I assume that seams192 wants to prove $\displaystyle |sin(x)- sin(y)|\le |x- y|$ in order to prove that sin(x) is continuous and you can't use the mean value theorem unless you know that sin(x) is continuous and differentiable.

If you define sin(t) as the y coordinate of the point (x, y) a distance t from (1, 0) on the unit circle (there are many different way to define sin(x) but seams192 specifically mentions the unit circle), the standard method is to the geometry of the circle. Look at the circular sector from (1, 0) to (cos(t), sin(t)). Since the definition of radian measure**is**"the arclength swept out on the unit circle", the length around the unit circle from (cos(t), sin(t)) to (1, 0) is t.

Now, draw a perpendicular from (x, y)= (cos t, sin t) to (cos(t), 0). Its length is sin(t). Further, the straight line from (cos t, sin t) to (1, 0) is shorter than the arc (since a straight line is the shortest distance between two points) so is less than t. But that straight line is the hypotenuse of the right triangle with vertices at (cos(t), sin(t)), (cos(t), 0), and (1, 0) while the line from (cos(t), sin(t)) to (cos(t), 0) is a leg of that triangle.

The hypotenuse of a right triangle is longer than any leg so we have

0< sin(t)< h< t where h is the length of the straight line from (cos t, sin t) to (1, 0).

Now take the limit as to goes to 0 and we have $\displaystyle \lim_{t\to 0} sin(t)= 0$. Since sin(0)= 0, that proves that sin(t) is continuous at t= 0.

Since $\displaystyle sin^2(t)+ cos^2(t)= 1$ for all t, taking limit as t goes to 0, we get $\displaystyle \lim_{t\to 0} cos^2(t)= 1$ and, since cos(t) is positve for all t near 0, $\displaystyle \lim_{t\to 0} cos(t)= 1$.

Now, let $\displaystyle x_0$ be any real number. Let $\displaystyle h= x- x_0$ where x can be any real number. Then $\displaystyle x= x_0+ h$ and $\displaystyle \lim_{x\to x_0} sin(x)= \lim_{h\to 0} sin(x_0+ h)$.

But $\displaystyle sin(x_0+ h)= sin(x_0)cos(h)+ cos(x_0)sin(h)$ so

$\displaystyle \lim_{x\to x_0} sin(x)= \lim_{h\to 0} sin(x_0+ h)$[/tex]= sin(x_0)\lim_{h\to 0}cos(h)+ cos(x_0)\lim{h\to 0}sin(h)= sin(x_0)(1)+ cos(x_0)(0)= sin(x_0)[/tex], showing that sin(x) is continuous for all x. - May 19th 2010, 07:07 AMDrexel28
It's true in fact that for any functions continuous $\displaystyle c(x),s(x)$ which are continuous at $\displaystyle 0$ and $\displaystyle c^2(x)+s^2(x)=1$ and $\displaystyle s(x+y)=c(x)s(y)+c(y)s(x)$, and $\displaystyle c(x+y)=c(x)c(y)-s(x)s(y)$ they are continuous. Can you see why OP?