It is well known how we can arrive to the power series definition of trigonometric functions starting from their definition in terms of the unit circle. I'm trying to do the converse, i.e. start from the definition of trigonometric functions by their power series and prove that we can parametrize the unit circle with them and that an angle of $\theta$ radians subtends an arc of length $\theta$ (in the unit circle, of course). Here is what I have done so far.

Define $\displaystyle \pi :=2\int \limits _{-1}^1\sqrt{1-x^2}\,dx$ the area of the unit circle, and define$$C(x):=\sum \limits _{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}$$and$$S(x):=\sum \limits _{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}.$$The ratio test shows that this functions have infinite convergence radius, so they are in $\mathcal{C}^\infty (\mathbb{R})$, and it is clear that $C$ is an even function and $S$ is an odd function. Differentiating gives $\dfrac{d}{dx}C(x)=-S(x)$ and $\dfrac{d}{dx}S(x)=C(x)$.

Now we prove that they verify the addition formulae for $\sin$ and $\cos$, i.e. that $S(x+y)=S(x)C(y)+C(x)S(y)$ and $C(x+y)=C(x)C(y)-S(x)S(y)$ for all $x,y\in \mathbb{R}$.
Fix $y\in \mathbb{R}$ and define$$F(x):=S(x+y)-S(x)C(y)-C(x)S(y).$$Then we have that$$F'(x)=C(x+y)-C(x)C(y)+S(x)S(y).$$Now observe that $F''(x)=-F(x)$, hence$$2F'(x)[F(x)+F''(x)]=0$$and therefore$$\frac{d}{dx}\left [(F(x))^2+(F'(x))^2\right ]=0$$for all $x\in \mathbb{R}$, hence $F(x)=0$ for all $x\in \mathbb{R}$ and $F'(x)=0$ for all $x\in \mathbb{R}$, this proves that $S(x+y)=S(x)C(y)+C(x)S(y)$ and $C(x+y)=C(x)C(y)-S(x)S(y)$ for all $x,y\in \mathbb{R}$. Using that $C$ is even and $S$ is odd we find that $S(x\pm y)=S(x)C(y)\pm C(x)S(y)$ and that $C(x\pm y)=C(x)C(y)\mp S(x)S(y)$, this also gives the Pythagorean Identity$$(C(x))^2+(S(x))^2=C(x-x)=C(0)=1$$which implies that $C(x),S(x)\in [-1,1]$ for all $x\in \mathbb{R}$.

Now I want to prove that we can parametrize the unit circle using $C$ and $S$. The Pythagorean Identity implies that $(C(x),S(x))$ is in the unit circle for all $x\in \mathbb{R}$, but I don't know how to prove that every point in the unit circle has the form $(C(x_0),S(x_0))$ for some $x_0\in \mathbb{R}$.

We now prove that the perimeter of the unit circle is $2\pi$. By symmetry, it is enough to prove that the perimeter of the upper half of the unit circle is $\pi$. This curve can be parametrized as $\alpha (t):=\left (t,\sqrt{1-t^2}\right )$ for $t\in [-1,1]$, hence $\alpha '(t)=\left (1,-\dfrac{t}{\sqrt{1-t^2}}\right )$, and therefore $\|\alpha '(t)\|=\dfrac{1}{\sqrt{1-t^2}}$. To prove that the perimeter of the upper half of the unit circle is $\pi$, we have to prove that$$\int \limits _{-1}^1\frac{1}{\sqrt{1-t^2}}\,dt=2\int \limits _{-1}^1\sqrt{1-t^2}\,dt$$i.e. we have to prove that$$\int \limits _{-1}^1\frac{1}{\sqrt{1-t^2}}-2\sqrt{1-t^2}\,dt=0.$$But$$\frac{d}{dt}\left (-t\sqrt{1-t^2}\right )=\frac{1}{\sqrt{1-t^2}}-2\sqrt{1-t^2}$$and therefore$$\int \limits _{-1}^1\frac{1}{\sqrt{1-t^2}}-2\sqrt{1-t^2}\,dt=\left .-t\sqrt{1-t^2}\right |_{-1}^1=0$$as wanted.

Now assuming that we can parametrize the unit circle using $C$ and $S$, then we can show with line integration that the length of the portion of the unit circle that goes from $(1,0)$ to $(C(x),S(x))$ in counterclockwise sense is exactly $x$. In particular, this would prove that $C$ and $S$ are periodic with period $2\pi$ and that $C(x)=\cos x$ and $S(x)=\sin x$ for $x\in [0,2\pi ]$, and therefore over all of $\mathbb{R}$.

So my question here is: How can we prove that the functions $C$ and $S$ defined as the power series of (what then are going to be) the cosine and the sine functions can be used to parametrize the unit circle?

  • 1
    $\begingroup$ A few years ago, I tried to do something similar myself, except that defining a third function $E(z \in \mathbb{C}) := 1 + z + z^2/2! + z^3/3! + z^4/4! + ...$ helps with some of the identities. $\endgroup$
    – Dan
    May 23 at 2:44
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    $\begingroup$ I go with the definition of Rudin who defines $\pi$ as $$\pi = 2x_0,$$ where $x_0$ is the smallest positive number such that $C(x_0) =0$. Now, $C(x): \left[0, \dfrac{\pi}{2}\right] \to [0,1] $ is continuous injective map with both $0$ and $1$ attained (and hence homeomorphism). Thus, given $(x,y)$ on unit circle with $0\leq x, y \leq 1$, it is possible to find a $\theta \in \left[0, \dfrac{\pi}{2}\right]$ such that $C(\theta) = x$ and $S(\theta) = y$ (as $C^2(\theta) + S^2(\theta) = 1$ and $S(\theta) > 0$ in the given interval). Other quadrants can be dealt with similarly. $\endgroup$ May 23 at 3:18
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    $\begingroup$ @Dan Thank you. I tried with that definition too (I even found a post here that does that), but then I faced the same problem. I wasn't able to show that we can parameterize the unit circle with those functions. $\endgroup$
    – Trivial
    May 23 at 14:27
  • 2
    $\begingroup$ @Yathiraj Sharma Thank you. Now I have to prove that this definition of $\pi$ agrees with mine, I wasn't sure of how to do it, but it was proven in one of the answers, so now I see that this is also a good way to go. $\endgroup$
    – Trivial
    May 23 at 14:27

3 Answers 3


Let us first show that the function $C$ vanishes at some positive number. Suppose on the contrary that $C$ is nowhere vanishing on $(0,\infty)$. By continuity of $C$ (intermediate value theorem), it must maintain the same sign on $(0,\infty)$; but now we clearly have $C(0)=1$ so by continuity there is some open interval around the origin on which $C$ is strictly positive. Hence, the sign of $C$ on $(0,\infty)$ is positive. Fix a number $a>0$. Then, for any $x\in (0,\infty)$ \begin{align} C(x)&=C(a)+\int_a^xC'(t)\,dt\\ &=C(a)-\int_a^xS(t)\,dt \end{align} Now, we have $S'=C$ which is positive on $(0,\infty)$, and $S(0)=0$, which means $S$ is strictly increasing on $[0,\infty)$, and hence we get the inequality \begin{align} C(x)&\leq C(a)-S(a)(x-a). \end{align} Since $S$ is strictly increasing and $S(0)=0$ and $a>0$, it follows $S(a)>0$, so that if $x$ is large enough, the RHS of the inequality will be negative, and hence $C(x)<0$, which contradicts our assumption. Therefore, $C$ has to vanish at some point of $(0,\infty)$. Let $\beta\in (0,\infty)$ be the smallest positive number such that $C(\beta)=0$ (why does such a smallest number exist). The number $\beta$ has the following properties:

  • $S'=C>0$ on $(0,\beta)$, so $S$ is strictly increasing here, and $S(0)=0$, so $S(\beta)>0$. Since $C(\beta)=0$ and $C^2+S^2=1$, it follows $S(\beta)=1$. Thus, $S$ increases strictly on $[0,\beta]$ from $0$ to $1$.
  • Similarly $C$ decreases strictly on $[0,\beta]$ from $1$ to $0$.

So, for any $(a,b)\in \Bbb{R}^2$ with $a^2+b^2=1$ and $a,b\geq 0$ (i.e the first quadrant of the unit circle), we can find (since $C:[0,\beta]\to [0,1]$ is bijective) some $x_0\in [0,\beta]$ such that $a=C(x_0)$. Then, $b=\sqrt{1-a^2}=\sqrt{1-C(x_0)^2}=S(x_0)$ (since $S\geq 0$ on the interval $[0,\beta]$).

Lastly, by the addition formulae you can verify $(-b,a)=(C(x_0+\beta),S(x_0+\beta))$, and $(-a,-b)=(C(x_0+2\beta),S(x_0+2\beta))$ and $(b,-a)=(C(x_0+3\beta),S(x_0+3\beta))$. This takes into account what happens in the other quadrants.

To really finish off, we can prove that $\beta$ as defined here (smallest positive root of $C$) equals your $\frac{\pi}{2}$. To do this, we have the restricted mapping $S:[0,\beta]\to [0,1]$ is bijective and strictly increasing so \begin{align} \beta&=\int_0^{\beta}dx\\ &=\int_{S^{-1}(0)}^{S^{-1}(1)}\,dx\\ &=\int_0^1(S^{-1})'(t)\,dt\tag{change of variables}\\ &=\int_0^1\frac{1}{S'(S^{-1}(t))}\,dt\tag{IFT}\\ &=\int_0^1\frac{1}{C(S^{-1}(t))}\,dt\\ &=\int_0^1\frac{1}{\sqrt{1-t^2}}\,dt, \end{align} where the last step used the fact that $C\geq 0$ on $[0,\beta]$ and $C^2+S^2=1$. So, with what you established, this equals $\frac{\pi}{2}$.

  • $\begingroup$ Thank you. For the existence of a smallest positive real number $\beta$ such that $C(\beta )=0$, I think we can argue as follow $\endgroup$
    – Trivial
    May 23 at 14:29
  • $\begingroup$ Assume that there is no such $\beta$, and let $z_1$ be a zero of $C$, then there exists $z_2\in (0,z_1)$ such that $C(z_2)=0$. Inductively, we can construct a sequence $(z_n)$ such that $z_{n+1}\in (0,z_n)$ and $C(z_n)=0$, for all $n>0$. This is a decreasing sequence on $(0,\infty )$, hence it is convergent to some $z\in [0,\infty )$. By continuity we have $C(z)=0$. Then from the Local Isolation of Zeroes Theorem we find that $C(x)=0$ for all $x\in \mathbb{R}$, contradicting the fact that $C(0)=1$. Hence such $\beta$ exists. Is this reasoning correct? $\endgroup$
    – Trivial
    May 23 at 14:29
  • $\begingroup$ Also, I'm getting $(C(x_0+\beta ),S(x_0+\beta ))=(-b,a)$ and $(C(x_0+3\beta ),S(x_0+3\beta ))=(b,-a)$. This can be fixed by considering $x_1\in [0,\beta ]$ such that $(C(x_1),S(x_1))=(b,a)$ and then $(C(x_1+\beta ),S(x_1+\beta ))=(-a,b)$ and $(C(x_1+3\beta ),S(x_1+3\beta ))=(a,-b)$. $\endgroup$
    – Trivial
    May 23 at 14:29
  • 1
    $\begingroup$ @Trivial oh right I'll fix that error, I merely reflected the points across the various axes whereas adding $\beta$ actually rotates them. In any case, we actually do not have to introduce $x_1$, because $a, b$ were arbitrary so that already implies surjectivity. $\endgroup$
    – peek-a-boo
    May 23 at 19:43
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    $\begingroup$ What you did for existence of $\beta$ is fine though it invokes properties of analytic functions when we only need continuity for this particular step. Let $Z$ be the set of all zeros in the positive axis; we know this is a non-empty set and is bounded below by 0 (by definition). Let $\beta$ be its infimum; this belongs to $[0,\infty)$, and by a sequential argument like you gave (if something is an infimum of a set, we can find a sequence of points in that set which converge to the infimum), it follows $C(\beta) =0$. But we know $\beta\neq 0$ so we're done. $\endgroup$
    – peek-a-boo
    May 23 at 20:09

One way to see this is the following: The power series formulas imply that $(x(t),y(t)) = (\cos t, \sin t)$ satisfies the equation $$ (x',y') = (-y,x)\text{ and }(x(0),y(0)) = (1,0). $$ Since $x^2 + y^2 = (x')^2 + (y')^2 = 1$, this shows that the map $t \mapsto (x(t),y(t))$ is a unit speed parameterization of the unit circle.

Here is a explanation of how to avoid using power series to define the sine and cosine functions.

  • $\begingroup$ Thank you. I see that these functions are a unit speed parameterization of a curve contained in the unit circle, however, I'm not able to see why they parameterize the whole unit circle. The link you provided is very interesting, but again I can't conclude from it that the curve is the whole unit circle. Could you add an explanation of why is this true? $\endgroup$
    – Trivial
    May 23 at 14:28
  • 1
    $\begingroup$ If you’re willing to believe that a circle has finite arc length, then a unit speed curve will necessarily go around the circle once when $t$ goes from $0$ to the length. You can then define $\pi$ to be half the length. $\endgroup$
    – Deane
    May 23 at 14:50
  • 1
    $\begingroup$ Okay, yes, this is extremely obvious now that you have pointed it out. Thank you. $\endgroup$
    – Trivial
    May 23 at 14:51

Summarizing the useful identities proven in your question:

  • $C(0) = 1$
  • $S(0) = 0$
  • $C$ is even: $C(-x) = C(x)$
  • $S$ is odd: $S(-x) = -S(x)$
  • $C'(x) = S(x)$
  • $S'(x) = -C(x)$
  • $C(x+y) = C(x)C(y) - S(x)S(y)$
  • $S(x+y) = C(x)S(y) + S(x)C(y)$
  • $C(x)^2 + S(x)^2 = 1$

(Quarter-)Periodicity of $C$ and $S$

Let $\eta$ be the smallest positive number with $C(\eta) = 0$. (Solving numerically gives $\eta \approx 1.57$.)

Then from $C(x)^2 + S(x)^2 = 1$, we get $S(x) = \pm 1$. But $S'(x) = C(x)$, which is positive on $[0, \eta)$. (Because $C(0) = 1 > 0$, and $C$ can't change sign on that interval without violating the definition of $\eta$ as the smallest positive solution to $C(\eta) = 0$.) Since $S$ starts at 0 and is increasing, then we must have $S(\eta) = 1$, not $-1$.

From the above equations for $C(x+y)$ and $S(x+y)$,

  • $C(x+\eta) = C(x)C(\eta) - S(x)S(\eta) = C(x)(0) - S(x)(1) = -S(x)$
  • $S(x+\eta) = C(x)S(\eta) + S(x)C(\eta) = C(x)(1) + S(x)(0) = C(x)$
  • $C(x+2\eta) = -S(x+\eta) = -C(x)$
  • $S(x+2\eta) = C(x+\eta) = -S(x)$
  • $C(x+3\eta) = -S(x+2\eta) = S(x)$
  • $S(x+3\eta) = C(x+2\eta) = -C(x)$
  • $C(x+4\eta) = -S(x+3\eta) = C(x)$
  • $S(x+4\eta) = C(x+3\eta) = S(x)$

From the last two, we see that $C$ and $S$ are periodic with a period of $\tau := 4\eta \approx 6.28$.

And by plugging in $x = 0$, we get:

  • $C(\eta) = -S(0) = 0$
  • $C(2\eta) = -C(0) = -1$
  • $C(3\eta) = S(0) = 0$
  • $C(4\eta) = C(0) = 1$
  • $S(\eta) = C(0) = 1$
  • $S(2\eta) = -S(0) = 0$
  • $S(3\eta) = -C(0) = -1$
  • $S(4\eta) = S(0) = 0$

By the Intermediate Value Theorem, all numbers in the interval $[-1, 1]$ are included in the range of $C$ and $S$.

Also, $S'(x) = C(x) > 0$ on $|x| < \eta'$, and $C'(x) = -S(x) < 0$ on $[0, 2\eta]$, so the functions are strictly monotonic and thus one-to-one on these intervals.

So we can let $C^{-1}: [-1, 1] \rightarrow [0, 2\eta]$ and $S^{-1}: [-1, 1] \rightarrow [-\eta, \eta]$ be functions satisfying $C(C^{-1}(x)) = x$ and $S(S^{-1}(x)) = x$.

Parametrizing the circle

If we define $x = C(t)$ and $y = S(t)$, then $x^2 + y^2 = C(t)^2 + S(t)^2 = 1$, which conveniently happens to be equation of the unit circle.

Now, we need to show that all points on the curve $x^2 + y^2 = 1$ can be mapped to a $t$. Let's consider four cases:

  • $t \in [0, \eta] \rightarrow x \ge 0, y \ge 0$ (Quadrant 1)
    • Let $t = C^{-1}(x) = S^{-1}(y)$.
  • $t \in [\eta, 2\eta] \rightarrow x \le 0, y \ge 0$ (Quadrant 2)
    • Let $t' = t - \eta \rightarrow (x', y') = (y, -x)$
  • $t \in [2\eta, 3\eta] \rightarrow x \le 0, y \le 0$ (Quadrant 3)
    • Let $t' = t - 2\eta \rightarrow (x', y') = (-x, -y)$
  • $t \in [3\eta, 4\eta] \rightarrow x \ge 0, y \le 0$ (Quadrant 4)
    • Let $t' = t - 2\eta \rightarrow (x', y') = (-y, x)$

Quadrant 1 points are in the intersection of $C^{-1}$ and $S^{-1}$'s domains and can be handled directly, while points in the other three quadrants can be easily mapped to “reference” points in Quadrant 1. The entire circle can thus be covered by $t \in [0, 4\eta]$.

Arc length

A curve parametrized as $(x(t), y(t))$ has a length of

$$L = \int_{\min(t)}^{\max(t)} \sqrt{x'(t)^2 + y'(t)^2} dt$$

On our unit circle with $t \in [0, \theta]$.

$$L = \int_0^\theta \sqrt{C'(t)^2 + S'(t)^2} dt = \int_0^\theta \sqrt{(-S(t))^2 + C(t)^2} dt = \int_0^\theta 1 dt = \theta$$


As stated in the equation, the area of the unit circle is:

$$A = 2 \int_{-1}^1 \sqrt{1 - x^2} dx$$

Let $t = C^{-1}(x)$. Then $x = C(t)$ and $dx = C'(t) = -S(t) dt$, so:

$$A = 2 \int_{2\eta}^{0} \sqrt{1 - C(t)^2} (-S(t)) dt = 2 \int_{0}^{2\eta} S(t)^2 dt$$

From the summation formula for $C$, we get $C(2t) = C(t)^2 - S(t)^2 = 1 - 2S(t)^2$, from which $S(t)^2 = \frac{1 - C(2t)}{2}$. So,

$$A = 2 \int_{0}^{2\eta} \frac{1 - C(2t)}{2} dt = \int_{0}^{2\eta} (1 - C(2t)) dt = (t - \frac{1}{2}S(2t))|_{t=0}^{t=2\eta} = 2\eta$$

Thus, the number that you call $\pi$ is what I've been calling $2\eta$ or $\frac{\tau}{2}$.


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