I'd like to revisit a question that has been brought up a couple of times here on math.stackexchange, yet as far as I'm aware, nobody has come up with a final answer.

The question is: Is it possible to give a rigorous definition of sine and cosine not using the power series approach?

One of my beginners courses at the university for example did like that:

  1. They gave us the intuitive definition of sine and cosine using the unit circle.

  2. The explicitly mentioned that these definitions were dependent on arc length, whose definition they would omit here though but rather rely on intuition.

  3. Then they said, the following properties could (in theory) be proved but to avoid the above mentioned challenges, they'd rather introduce them as axioms:

    1. Sine and cosine are defined on the real domain and are continuous.

    2. Sine is an odd and cosine an even function.

    3. They would abide to the respective addition formulas.

    4. $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$.

    5. $\cos(0) = 1$.

  4. They then state that these five axioms already uniquely define a pair of functions. In other words, if a pair of functions are subject to the five axioms, they are the sine and the cosine.

  5. Eventually, the introduce the well-known power series for both sine and cosine and show that they abide to the axioms and hence we know, they are the sine and the cosine.

When you try to prove the five "axioms" then all difficulties are related to the arc length. The usual way to introduce the arc length would require me to know the derivate of sine and cosine but using that is prohibitive here as we have not yet defined sine and cosine.

To cut a long story short: Can be done and has it been done by someone? A purely geometrical yet rigorous definition of sine and cosine or can it proved that it's doomed to fail as the circular dependencies can't be avoided?

Looking forward to your feedback!

  • $\begingroup$ The five axioms in 3 do not look geometric $\endgroup$
    – Henry
    Apr 20, 2023 at 9:27
  • $\begingroup$ Related - math.stackexchange.com/questions/1078377/… $\endgroup$
    – Henry
    Apr 20, 2023 at 9:36
  • $\begingroup$ Partly true (some are, some not). But the idea is to come up with a geometric definition that allows to derive them. IMHO that should work the moment you have a definition of arc length that doesn't depend on anything you want to prove. $\endgroup$ Apr 20, 2023 at 9:44
  • 1
    $\begingroup$ You don't need the arc length, you can find the limit of $\sin(x)/x$ with the area of a circular sector and two triangles. There is still some work to make it rigorous enough. $\endgroup$ Apr 20, 2023 at 9:51
  • 2
    $\begingroup$ In the title you ask about a geometric definition whereas in the body there is no mention about geometry. $\endgroup$
    – user
    Apr 20, 2023 at 16:30

2 Answers 2


We can use the exponential map between the Lie algebra and the Lie group to provide a rigorous definition of sine and cosine. Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$, and let $X\in \mathfrak{g}$. The exponential map takes an element of $\mathfrak{g}$ and maps it to an element of $G.$

Let $H$ be a closed subgroup of $G$, and let $\mathfrak{h}$ be its Lie algebra. We define the orthogonal complement of $\mathfrak{h}$ in $\mathfrak{g}$ as $\mathfrak{h}^{\perp}$, which is the set of all elements of $\mathfrak{g}$ that are orthogonal to $\mathfrak{h}$ with respect to the Killing form of $\mathfrak{g}$.

We can then define the sine and cosine functions as follows: \begin{align*} \cos(X) & = \mathrm{proj}H(\exp(X))\\ \sin(X) & = \mathrm{proj}\mathfrak{g}^{\perp}(\exp(X)) \end{align*} Here, $\mathrm{proj}H$ and $\mathrm{proj}\mathfrak{g}^{\perp}$ are the orthogonal projections of $\exp(X)$ onto $H$ and $\mathfrak{g}^{\perp}$, respectively.

To show that these definitions are rigorous, we need to show that they satisfy the familiar properties of cosine and sine functions. One way to do this is to use the properties of the exponential map, which include:

  • $\exp(0) = e$ (the identity element of $G$)
  • $\frac{d}{dt}(\exp(tX)) = X\exp(tX)$
  • $\exp(X)\exp(Y) = \exp(X+Y)$ (provided $[X,Y]=0$)

Using these properties, we can show that:

  1. $\cos(0) = \mathrm{proj}H(\exp(0)) = e$ (the identity element of $H$)
  2. $\frac{d}{dt}(\cos(tX)) = -\sin(tX)$
  3. $\cos(X)\cos(Y) = \mathrm{proj}H(\exp(X))\mathrm{proj}H(\exp(Y)) = \mathrm{proj}H(\exp(X)\exp(Y)) = \mathrm{proj}H(\exp(X+Y)) = \cos(X+Y)$

Similarly, we can show that:

  1. $\sin(0) = \mathrm{proj}H^{\perp}(\exp(0)) = 0$ (the zero element of h^⊥)
  2. $\frac{d}{dt}(\sin(tX)) = \cos(tX)$
  3. $\sin(X)\sin(Y) = \mathrm{proj}H^{\perp}(\exp(X)) \mathrm{proj}H^{\perp}(\exp(Y)) = \mathrm{proj}H^{\perp}(\exp(X)\exp(Y)) = \mathrm{proj}H^{\perp}(\exp(X+Y)) = \sin(X+Y).$
  • $\begingroup$ That's a rather interesting approach and the first time I heard of Lie groups in the context of trigonometric functions. You deserve credits just for the - from my point of view - novel approach. However, doesn't its reliance on the exponential function introduce power series via the backdoor? $\endgroup$ Apr 20, 2023 at 11:10
  • 2
    $\begingroup$ Usually it is $\exp(1)=e$... (just kidding !) $\endgroup$
    – Jean Marie
    Apr 20, 2023 at 12:00

I have never seen a book using a purely geometric definition but it can be done. I will try to give you the outline of such an approach and if you wish I can send you a pdf file I am currently writing for the details.

Actually in some way the rigorous approach is just the intuitive approach, so let us take a look at the intuitive approach and try to fix it's problems. To get $(\cos(x), \sin(x))$ by the intuitive approach lets proceed as follows: Take the point $z$ on the circle which is at an arc length of $x$ from $(1, 0)$ in the direct orientation and then define $\cos(x) = p_1(z)$ and $\sin(x) = p_2(z)$ where $p_1$ and $p_2$ are the projections.

The problem of this intuitive approach is that many of the terms used are undefined and it is unclear if such a point $z$ exists or is unique. The first step is to define what is arc length, to do so we define what is the length of a curve this is done here: https://en.wikipedia.org/wiki/Arc_length. The second step would be to define what it means to be directly oriented, to avoid using differential geometry we can use this simple definition: a differentiable curve $\gamma \in \mathcal{C}([a, b], \mathcal{S}^1)$ is said to be directly oriented if: $\forall t \in [a, b], \det(\gamma(t), \gamma'(t)) > 0$. Now suppose that there is $\gamma \in \mathcal{C}([a, b], \mathcal{S}^1)$ such that the following are true:

  1. $\gamma$ is $\mathcal{C}^1$ by parts.
  2. $\gamma([a, b]) = \mathcal{S}^1$
  3. $\gamma$ is a simple loop.
  4. $\gamma$ is directly oriented.
  5. $\gamma(a) = (1, 0)$

Using such a curve $\gamma$ we can now define the functions $\cos$ and $\sin$ and we shall proceed by keeping the idea of the intuitive definition. First given an $x \in [a, b]$ we should find a point $z$ on the circle which is at an arc length of $x$ from $(1, 0)$ in the direct orientation. To do so we can consider the function $f(x) = Length(\gamma \lvert_{[a, x]}) = \underset{[a, x]}{\int} \lVert \gamma'(t) \rVert d \lambda(t)$ and notice that singe $\gamma$ is directly oriented $\gamma'(x) \neq 0$ thus $f'(x) = \lVert \gamma'(x) \rVert > 0$ means that $f$ is a diffeomorphism. Thus the point $z$ would be given by $z = \gamma(f^{-1}(x))$ and now by using the projections $p_i$ we can define: $$\cos(x) = p_1 \circ \gamma \circ f^{-1}(x)$$ $$\sin(x) = p_2 \circ \gamma \circ f^{-1}(x)$$ We can also prove this definition is independent of $\gamma$. And you may ask that here we used the fact that such a $\gamma$ exists but how do we prove it? To do so we may just use the equation $x^2 + y^2 = 1$ and obtain curves that look like $\alpha(t) = \left(t , (1 - t^2)^{\frac{1}{2}} \right)$ and them "stick" them together to obtain the wanted curve.

This definition is very close to the intuitive definition however to prove the theorems you need something a bit better, bellow I will offer an other definition that still keeps this geometric spirit but is easier to use for proving things. It is based on the following result: For every $z \in \mathcal{S}^1$ there is a unique curve that we will call $trig_z \in \mathcal{C}([0, a], \mathcal{S}^1)$ which verifies all the following:

  1. $trig_z$ is $\mathcal{C}^1$ and is a simple loop
  2. $trig_z(0) = z$
  3. $trig_z([0, a]) = \mathcal{S}^1$
  4. $trig_z$ is directly oriented
  5. $\lVert trig_z'(x) \rVert = 1$

Of course to prove the previous result no trigonometry is needed but the proof is not so simple. After by defining $trig = trig_{(1, 0)}$ then we can define: $$\cos = p_1 \circ trig$$ $$\sin = p_2 \circ trig$$

Using $trig$ you can prove all the results, for example to get the differentials of $\cos$ and $\sin$ you can differentiate the equation: $$\cos(x)^2 + \sin(x)^2 = 1$$ Or use the fact that $trig'(x)$ must be in the tangent line $T_x \mathcal{S}^1$ of $\mathcal{S}^1$.

You can also prove the $\cos(x + y)$ and $\sin(x + y)$ formulas by defining $\gamma(y) = \cos(y) trig(x) + \sin(y) trig'(x)$ we can check that $\gamma(0) = trig(x)$ and $\gamma$ has all the required properties to apply our uniqueness result and we obtain that $\gamma = trig_{trig(x)}$. And we can also show that $trig_{trig(x)} = trig(\cdot + x)$. Thus by noticing that $trig'(x) = (- \sin(x), \cos(x))$ and using that $\gamma = trig{\cdot + x}$ we obtain the wanted formulas.

Using such techniques you can prove everything, well at least for everything I tried to prove, I was able to.

I hope this replies to your question and helped you.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .