Proof that $(\sin(x),\cos(x))$ describe a circle? In my analytics class $\sin$ and $\cos$ were defined as follows:
$$ \sin(x) =  \sum_{n=0}^\infty \frac{(-1)^{n}x^{2n+1}}{(2n+1)!} \text{ and } \cos(x) =  \sum_{n=0}^\infty \frac{(-1)^{n}x^{2n}}{(2n)!}$$ 
But how does one prove that $\sin$ and $\cos$ are actually the functions that are defined by triangles? In particular I  want to prove that
$$t \to (\sin(t),\cos(t))$$ 
describes a circle for $t$ between $0$ and $2\pi$ where $\pi$ is defined as below. 

Here are things that have been proven in the class:
Also it was proven that
$$ \sin(x)^2  + \cos(x)^2 = 1$$
And a lot of formulas like this:


*

*$ \cos(x)=\cos(-x) \text{ and } \sin(x)=-\sin(-z)$

*$\sin(x+y) = \sin(x) \cos(y) + \sin(x)\cos(y)$
Also it was shown that $\cos(x)$ has a zero point in [0,2] which we  defined as $\frac{\pi}{2} $. Then we showed:


*

*$\sin(\pi)=0, \, \cos(\pi)=-1$

*$\cos(\pi+x) = -\cos(x), \, \sin(\pi+x) = -\sin(x)$

*$\sin(\frac{\pi}{2} + x) = \cos(x)$
 A: This probably isn't the best proof but it should work. First use $\sin^2(t)+\cos^2(t)=1$ to conclude that $(\cos(t),\sin(t))$ lies on the unit circle. Then, $(\cos(t),\sin(t))$ is continuous so it maps connected sets in $\mathbb{R}$ to connected sets in $\mathbb{R}^2$. In particular, the image of $[0,\pi]$ under $(\cos(t),\sin(t))$ is a connected set which lies on the unit circle and includes $(1,0)$ and $(-1,0)$. It follows that you must have either the upper or lower half circle in the image of $[0,\pi]$. 
Finally use $\cos(x+\pi)=-\cos(x)$ and $\sin(x+\pi)=-\sin(x)$ to conclude the image of $[0,2\pi]$ under $(\cos(t),\sin(t))$ is the unit circle.
A: This was a bit more complicated than I anticipated, but here is a complete proof.
A circle with radius $r$ is defined as the set of points in the plane at distance $r$ from the origin. The distance $d$ from the origin to a point $(x,y)$ is defined by 
$$ d = \sqrt{x^2+y^2} \implies d^2 = x^2 + y^2.$$
That is, the points $(x,y)$ on the circle with radius $r$ are precisely those that satisfy
$$ r^2 = x^2 + y^2.$$
If $r = 1$, this reduces to
$$ 1 = x^2 + y^2.$$
Hence the formula
$$ \cos^2(t) + \sin^2(t) = 1$$
is precisely the statement that $(\cos(t), \sin(t))$ lies on the unit circle. 
To show that every point on the unit circle is $(\cos(t), \sin(t))$ for some $t$, you can use the continuity of sine and cosine.
Let $(x,y)$ be a point on the unit circle. Since $x^2 + y^2 = 1$, the number $x$ lies in the interval $[-1,1]$. Since $\cos(0) = 1$ and $\cos(\pi) = -1$, there is some $s$ in the interval $[0,\pi]$ such that $\cos(s) = x$, via the intermediate value theorem.
Moreover
$$ x^2 + y^2 = 1$$
implies that
$$ y = \pm \sqrt{1-x^2} = \pm \sqrt{1-\cos^2(s)} = \pm \sqrt{\sin^2(s)} = \pm \sin(s).$$
If $y = \sin(s)$, then $(x,y) = (\cos(s),\sin(s))$. If $y = -\sin(s)$, then
$$ (x,y) = (\cos(s),-\sin(s)) = (\cos(-s), \sin(-s))$$
since sine is odd and cosine is even. Either way, the proof is complete.
A: In this post we give the characterization of monotony of $\sin $.
Put
$$
f(x)=\sum_{n=0}^\infty\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}
$$
and
$$
g(x) =\sum_{n=0}^\infty\frac{(-1)^{n}x^{2n}}{(2n)!}
$$
Obviously $f'(x)=g(x)$ and $g'(x)=-f(x)$.

Claim 1: $g$ has a minimum positive zero point $\pi/2$. 

Proof: Suppose for contradiction that $g$ has no positive zero point.
Considering that $g$ is continuous, from $g(0)=1$ we know $g(x)>0,\, \forall x>0$.
Thus $f$ is strictly increasing on $[0,\infty)$. Condering that $f$ is bounded, thus $\displaystyle\lim_{x \rightarrow +\infty} f(x)$ exists.
Denote $\displaystyle\lim_{x \rightarrow +\infty} f(x)= A$ where $A$ is a positive number.
Thus there exists a number $x_0$ s.t. $f(x)\ge \frac{A}{2}$ for all $x\ge x_0$.
On the other hand,
$$
g(x)=g(x_0)-\int_{x_0}^{x} f(t)dt \le g(x_0) -\frac{A}{2}(x-x_0), \, \forall x\ge x_0
$$
which contradicts that $g$ is bounded.
Thus $g$ has positive zero points. What's more, because $g$ is continuous , $g$ must have its minimum positive zero point. 
Just as you pointed out , the minimum positive zero point is $\pi/2$.

Claim 2: $f$ is strictly increasing on $[0,\pi/2]$

Proof: From $f^2+g^2=1$ we get $f(\pi/2)=1.$
Note that the solution for
$$
g''(x)=-g(x),\, g(0)=1,\, g'(0)=0 \tag{1}
$$
is unique.
Thus $f(\pi/2-x)$ , which also satisfies $(1)$, is equal to $g(x)$. i.e. $f(\pi/2-x)=g(x),\, \forall x\in\mathbb{R}$ 
Considering that $f$ is an odd function and $g$ is an even function, we get
$$
f(\pi-x)=f(\pi/2-(x-\pi/2))=g(x-\pi/2)=g(\pi/2-x)=f(x)
$$ 
Thus $f$ is strictly decreasing on $[\pi/2,\pi]$, which leads to claim 2.

Claim 3: $f$ is a function with period $2\pi$. And $f$ is strictly increasing on $[-\pi/2,\pi/2]$ and strictly decreasing on $[\pi/2,3\pi/2]$.

Considering $f(\pi+x)=f(-x)=-f(x)$  and all the materials above, it's easy to prove.
Combining $f^2+g^2=1$ , I think you can see that $\sin $ and $\cos $ describe the circle well.
A: Let's once and for all define $$\gamma(t)=(\cos(t),\sin(t)),$$where $\sin$ and $\cos$ are defined by those power  series.
You already know that $\gamma(t)$ lies on the unit circle, and it's already been pointed out that considerations of continuity and connectedness show that every point on the unit circle is  $\gamma(t)$ for some $t$. To show that $\sin$ and $\cos$ are the same as the functions defined in terms of triangles:
Considering only points in the first quadrant for simplicity.  If you think about the definition of radians in terms of arc length you see you need to prove this:


Suppose $0<x<\pi/2$. Let $p$ be the point on the unit circle in the first quadrant such that the arc along the circle (in the first quadrant) from $(1,0)$ to $p$ has length $x$. Then $p=\gamma(x)$.


Turning things around a bit, that's the same as this:


Suppose $0<x<\pi/2$ and let $p=\gamma(x)$. Then the arc on the unit circle (in the first quadrant) from $(1,0)$ to $p$ has length $x$.


And that's just calculus: $\sin^2+\cos^2=1$ shows that $|\gamma'(t)|=1$, so the arc length in question is $$\int_0^x|\gamma'(t)|\,dt=x.$$
A different argument - more complicated and actually less elementary: Say $S$ and $C$ are the functions defined by the "geometric" definitions of $\sin$ and $\cos$. In any calculus book you find a proof that $S'=C$ and $C'=-S$, in varying degrees of rigor. Hence $S$ and $\sin$ are both solutions to the IVP $$y''+y=0,\quad y(0)=0,y'(0)=1.$$So uniqueness for IVPs shows that $S=\sin$.
Heh: The second argument is definitely harder, since it requires a "geometric" proof that $S'=C$ and $C'=-S$. But you don't actually need any theorems about differential equations; uniqueness for the IVP in question is easy by a cute trick.
By linearity you need this:


If $y''+y=0$ and $y(0)=y'(0)=0$ then $y(t)=0$.


Proof. (WLOG $y$ is  real-valued.) Define $$f(t)=y(t)^2+(y'(t))^2.$$Since $y''+y=0$it follows that $f'=0$, so $f(t)=f(0)=0$. Since $y$ is real-valued, $y^2+(y')^2=0$ implies $y=0$.
