Proof that $(\sin(x),\cos(x))$ describe a circle?

Question

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:

• $\sin(\pi)=0, \, \cos(\pi)=-1$
• $\cos(\pi+x) = -\cos(x), \, \sin(\pi+x) = -\sin(x)$
• $\sin(\frac{\pi}{2} + x) = \cos(x)$

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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 $y=0$.