We all get familiar with elementary functions in high-school or college. However, as the system of learning is not that much integrated we have learned them in different ways and the connections between these ways are not clarified mostly by teachers. Once I read the calculus book by Apostol, I just found out that one can define these functions in a treatise systematic way only analytically. The approach used in the book with some minor changes is like this

$1.$ Firstly, introduce the natural logarithm function by $\ln(x)=\int_{1}^{x}\frac{1}{t}dt$ for $x>0$. Accordingly, one defines the logarithm function by $\log_{b}x=\frac{\ln(x)}{\ln(b)}$ for $b>0$, $b \ne 1$ and $x>0$.

$2.$ Then introduce the natural exponential function as the inverse of natural logarithm $\exp(x)=\ln^{-1}(x)$. Afterwards, introduce the exponential function $a^x=\exp(x\ln(a))$ for $a>0$ and real $x$. Interchanging $x$ and $a$, one can introduce the power function $x^a=\exp(a\ln(x))$ for $x \gt 0$ and real $a$.

$3.$ Next, define hyperbolic functions $\cosh(x)$ and $\sinh(x)$ by using exponential function

$$\matrix{ {\cosh (x) = {{\exp (x) + \exp ( - x)} \over 2}} \hfill & {\sinh (x) = {{\exp (x) - \exp ( - x)} \over 2}} \hfill \cr } $$

and then defining the other hyperbolic functions. Consequently, one can define the inverse-hyperbolic functions.

$4.$ Finally, the author gives three ways for introducing the trigonometric functions.

$\qquad 4.1-$ Introduces the $\sin x$ and $\cos x$ functions by the following properties

\begin{align*}{} \text{(a)}\,\,& \text{The domain of $\sin x$ and $\cos x$ is $\mathbb R$} \\ \text{(b)}\,\,& \cos 0 = \sin \frac{\pi}{2}=0,\, \cos \pi=-1 \\ \text{(c)}\,\,& \cos (y-x)= \cos y \cos x + \sin y \sin x \\ \text{(d)}\,\,& \text{For $0 \le x \le \frac{\pi}{2}$ we have $0 \le \cos x \le \frac{\sin x}{x} \le \frac{1}{\cos x}$} \end{align*}

$\qquad 4.2-$ Using formal geometric definitions employing the unit circle.

$\qquad 4.3-$ Introducing $\sin x$ and $\cos x$ functions by their Taylor series.

and then defining the other trigonometric ones and the inverse-trigonometric functions.

In my point of view, the approach is good but it seems a little disconnected as the relation between the trigonometric and exponential functions is not illustrated as the author insisted to stay in the real domain when introducing these functions. Also, exponential and power functions are just defined for positive real numbers $a$ and $x$ while they can be extended to negative ones.


$1.$ How many other approaches are used for this purpose? Are there many or just a few? Is there some list for this?

$2.$ Would you please explain just one of the other heuristic ways to introduce the elementary functions analytically with appropriate details?


  • Historical remarks are welcome as they provide a good motivation.

  • Answers which connect more advanced (not too elementary) mathematical concepts to the development of elementary functions are really welcome. As nice example of this is the answer by Aloizio Macedo given below.

  • It is hard to choose the best answer between these nice answers so I decided to choose none. I just gave the bounties to the ones that are more compatible with the studies from high-school. However, please feel free to add new answers including your own ideas or what you may think that is interesting so we can have a valuable list of different approaches recorded here. This can serve as a nice guide for future readers.

Useful Links

  • Here is a link to a paper by W. F. Eberlein suggested in the comments. The paper deals with introducing the trigonometric functions in a systematic way.

  • There are six pdfs created by Paramanand Singh who has an answer below. It discusses some approaches for introducing logarithmic, exponential and circular functions. I have combined them all into one pdf which can be downloaded from here. I am sure that it will be useful.

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    $\begingroup$ I've seen two similar approaches which both start with $\exp$. One considers the ODE $y'=y$, the other starts from the power series right off the bat. $\endgroup$
    – Ian
    Commented Dec 3, 2015 at 21:35
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    $\begingroup$ (1)If you define in and cos by power series it will seem strange and lacking motivation to students not conversant with trigonometry. They will not know why these particular power series should merit special interest. And you will then have to derive the geometric properties of sin and cos, including the angle-sum formulas, from the power series....(2). More in the style of exp and log , take sin and cos to be solutions of $f''=-f$ with (f(0),f'(0) equal (0,1) or (1,0). $\endgroup$ Commented Jan 1, 2016 at 14:30
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    $\begingroup$ For circular functions (i.e $\sin$ , $\cos$), I recommend reading this article "W. F. Eberlein Mathematics Magazine Vol. 39, No. 4 (Sep., 1966), pp. 197-201". you can read this online for free at jstor. $\endgroup$ Commented Jan 4, 2016 at 15:30
  • $\begingroup$ @user2838619: Thanks for the valuable reference. Can you kindly summarize the main points in the article and write it as an answer so it could be recorded in this thread? :) $\endgroup$ Commented Jan 4, 2016 at 15:42
  • $\begingroup$ sorry H.R. , but I can't summarize it in a good way. Anyway, for (5-A), knowing $\sin \frac{\pi}{2} = 1$ and $\sin x \ge 0$ for $0 \le x \le \frac{\pi}{2}$ will be enough for finding other properties of $\sin$ and $\cos$ $\endgroup$ Commented Jan 4, 2016 at 16:33

10 Answers 10


There are two canonical group structures in $\mathbb{R}$: $(\mathbb{R},+)$ and $(\mathbb{R}_{>0}, \cdot)$.

We search for the isomorphisms between the structures.

The identity is an automorphism on $(\mathbb{R},+)$ and the exponential is an isomorphism from $(\mathbb{R},+)$ to $(\mathbb{R}_{>0}, \cdot)$.

Furthermore, they are the only continuous such isomorphisms, once you fix a value on $1$.

So, we get:

The identity $id$ is the only continuous automorphism on $(\mathbb{R},+)$ such that $id(1)=1$ and the exponential $\exp$ is the only continuous isomorphism from $(\mathbb{R},+)$ to $(\mathbb{R}_{>0}, \cdot)$ such that $\exp(1)=e$.

From these, all other elementary functions follow.

Summarizing, in order to obtain the elementary functions, you only need the algebraically (and analytic, since we must suppose continuity) interesting ones.

Expanding a bit, if you don't want to be allowed to consider exponentiation to complex numbers, reaching $\sin$ and $\cos$ from $\exp$ and the identity may be troublesome. I will therefore provide another way of introducing $\sin$ and $\cos$. Ironically, it involves "complex" ideas.

Consider $C^{\infty}(\mathbb{R})$, and $X: C^{\infty}(\mathbb{R}) \rightarrow C^{\infty}(\mathbb{R})$ given by $$f \mapsto f'.$$ Consider also the identity function $I$ on $C^{\infty}(\mathbb{R})$. We have that $e^{x}$ and $e^{-x}$ are the two "moral" solutions (more precisely, they form a basis for the solutions) of $$X^2-I=0.$$ It is natural to search for the solutions of $$X^2+I=0.$$ (Seems familiar?) We then have that the solutions with appropriate initial conditions are $\sin$ and $\cos$.

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    $\begingroup$ This is important. It's not often stressed what the elementary functions represent in terms of algebraic structure. $\endgroup$ Commented Jan 2, 2016 at 16:28
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    $\begingroup$ Since the OP focuses on real scalars and even points out the restriction to the real domain causing 'disconnectedness between exponential and trigonometric functions' after point 5, you need to explain how "all other elementary functions follow" from the exponential in the context of real analysis. Also remember that the OP is interested in introducing the elementary functions, presumably to students who are taking a first rigorous course in real analysis. $\endgroup$
    – guest
    Commented Jan 3, 2016 at 20:20
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    $\begingroup$ I think you misinterpreted OP. Taking the whole sentence he wrote, on context: "In my point of view, the approach is good but it seems a little disconnected as the relation between the trigonometric and exponential functions is not illustrated as the author insisted to stay in the real domain when introducing these functions". He seems to be referring to the attempts and restrictions of the author mentioned as a hindrance, not as something he wants in an answer. $\endgroup$
    – Aloizio Macedo
    Commented Jan 3, 2016 at 20:25
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    $\begingroup$ Where you write $X$ at the end, most people would write $D$ (though I see what you're trying to draw a parallel to). $\endgroup$ Commented Jan 5, 2016 at 23:43
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    $\begingroup$ I think it's a bit circular to define $\textrm{exp}$ such that $\textrm{exp}(1)=e$, as $e$ is usually defined in terms of $\textrm{exp}$. $\endgroup$ Commented Jan 7, 2016 at 4:47

$1.$ Napier got approximate logarithms by using repeated squaring to compute, for example, that $(1.000001)^{693417}$ is about $2$. So $\log_{1.000001}2$ is about $693147.$ He would "normalize" logs to base $1+1/n$ by dividing them by $n$. The number we call $e$ kept showing up with a normalized log of approximately $1$. Thus the motivation for defining

$$\exp (x)=\lim_{n\to \infty}\left(1+\frac{x}{n}\right)^n$$

which is valid for all complex $x$.

$2.$ I have a fondness for defining $\log x=\int_1^x t^{-1}dt$ because it is so easy , by a linear change of variable, to show $\log a b =\log a+\log b$.

$3.$ H.Dorrie, in $101$ Great Problems In Elementary Mathematics, gives a short and simple deduction of the power series for sin and cos (given only $\sin'=\cos$ and $\cos'=-\sin$, and $x>0\to x>\sin x$, and that $\cos 0=1,\sin 0=0$ ) that requires no background in the general theory of power series, not even "finite power series plus remainder term."

  • $\begingroup$ @RonGordon: Yeah! and I gave the first bounty to this answer due to nice historical remarks. :) $\endgroup$ Commented Jan 7, 2016 at 14:11
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    $\begingroup$ First bounty for me....didn't expect it at all..... learning some of the history can help to understand the math : how it developed, what were the motivations, how the different math specialties have interacted, how it related to other sciences . This can help to develop a better understanding of the math itself. ..and it no longer seems like a collage of disconnected topics. $\endgroup$ Commented Jan 7, 2016 at 16:04
  • $\begingroup$ Moise's calculus book is work a good look at. $\endgroup$ Commented Jun 3, 2017 at 6:41

You can define $\sin$ and $\cos$ as solutions to the equation


The function $\sin$ is the unique solution satisfying $f(0) = 0, f'(0) = 1$, and $\cos$ is the unique solution satisfying $f(0) = 1, f'(0) = 0$. In other words, $\sin$ and $\cos$ are the functions describing the orbits of simple harmonic oscillators. We can then define $\pi$ as the half-period of $\sin$ (once we prove that it's periodic). In other words, $\pi$ is the time taken for a harmonic oscillator to go from one extreme value to the other (and therefore really has nothing to do with circles).

Now let $(x(t), y(t))$ be the coordinates of a particle moving around a unit circle at uniform speed $1$. Since the distance of the particle to the origin is constant, the velocity vector $(x(t), y(t))'$ must be orthogonal to $(x(t), y(t))$, and it's of unit length since the particle has unit speed. Therefore $(x(t), y(t))'=(-y(t), x(t))$.

From this equation we deduce $x''=-x$ and $y''=-y$, and the initial conditions are fixed by our assumptions about the nature of the circular motion. In other words, we've shown that a particle moving uniformly in a circle is a simple harmonic oscillator. Therefore $x=\sin, y=\cos$ (and therefore the time taken for one rotation is $2\pi$, whence the perimeter formula).


I quite like the approach, taken in some Russian and Bulgarian books, e.g Fundamentals of mathematical analysis by V.A. Ilyin and E.G. Poznyak and Mathematical Anlaysis by Ilin, Sadovnichi and Sendov. The benefits of the approach is that they use continuity, monotonicity and (mostly) elementary concepts, which students should know from high school.

We start with the exponential function. For $a > 0 \text{ and } x = \frac{p}{q} \in \mathbb{Q} $ we know what is $a^x = a^\frac{p}{q} $ (of course, we should have proven the existence of $n$-th root already). Now prove this function (for now defined on the rational numbers only) is monotonic. At the end of the day, for $x \in \mathbb{R} $ define $a^x$ as the unique number $y$ with the following property: for all $ \alpha < x < \beta, \alpha, \beta \in \mathbb{Q}$ we have $a^\alpha \leq y \leq a^ \beta$. In other words, $a^x$ is defined by "extending via monotonicity".

Now we can define $\log_a (x)$ as the inverse function of $a^x$, that is: the number $t$ such that $a^t = x$ This is exactly the definition students should have been given in high school, so it should come as no surprise.

Next follow the trigonometric functions: in high school they are often defined like that. To make the definition rigorous we can use functional equations, in a manner similar to what OP wrote. A student should already know the $\sin(\alpha + \beta)$, $\cos(\alpha + \beta)$ formulae, and $\sin^2 x + \cos^2 x = 1$ so it should be fairly easy to comprehend that these properties sort of define $\sin$ and $\cos$. The definition is: There exists an unique pair of functions $f$ and $g$, defined over the real numbers, and satisfying the following conditions:

$1.$ $f(\alpha + \beta) = f(\alpha)g(\beta) + f(\beta) g(\alpha)$
$2.$ $g(\alpha + \beta) = g(\alpha)g(\beta) - f(\alpha)f(\beta) $
$3.$ $f^2(x) + g^2(x) = 1$
$4.$ $f(0) = 0 , g(0) = 1, f(\frac{\pi}{2}) = 1, g(\frac{\pi}{2}) = 0$

We define $\sin(x) = f(x) $ and $\cos(x) = g(x)$ After that, we can establish the known properties of the trigonometric functions and find their Taylor series. At the end, one notices the relation $e^{ix} = \cos(x) + i \sin(x)$

The number $e$

We define the number $e:= \lim_{n\to \infty} (1 + \frac{1}{n})^n$. After the definition of $a^x$ for real $x$ we can show that $\lim _{h \to 0} (1 + h)^{\frac{1}{h}} = e$. When we try to find the deriative of $\log_a(x)$ we will get: $$[\log_a(x)]' = \lim_{h \to 0} \frac{1}{x} \log_a \left( 1 + \frac{h}{x}\right)^\frac{x}{h}$$ By the continuity of the logarithm and the above limit we get $[\log_a(x)]' = \frac{\log_a (e)}{x}$. Thus, the natural base for the $\log$ is $e$. Because $a^x$ is the inverse of $\log_a(x)$ it's a simple calculation to show that $(a^x)' = a^x \log_e (a)$, and therefore the natural choice of the number $a$ is $e$.

Remarks: The above definitions use only continuity and monotonicity, no derivatives and integrals. For this reason, they are (arguably) more natural than definitions via differential equations: I highly doubt there is a student who has good intuition for the differential equation $f' = f$ but doesn't have an idea what is $a^x$. The main disadvantage of this approach is the length: it takes around $15$ pages without the proof of the existence of $\sin$ and $\cos$, and the proof itself is around $10$ pages more.


Not long enough or detailed enough for the bounty, and essentially a gloss on @Ian 's answer, but perhaps worth adding to the discussion.

Most students learn the trig functions in high school, and perhaps the exponential. I like to reintroduce the exponential in calculus as the function that's its own derivative, since the most important use for that function in applications is solving the differential equation $f'(x) = kf(x)$. The level of rigor in the definition depends on the overall level of rigor in the course.

Then the (natural) logarithm is the inverse function.

When you get to power series you connect the exponential and trig functions by deriving the identity $$ e^{ix} = \cos{x} + i \sin{x}. $$

Then you can define the hyperbolic functions with the analogous formula.


There are many ways to define the elementary functions (exponential, logarithmic and circular functions). I have treated the case of exponential and logarithmic functions in my blog posts:

All these approaches define the functions for real variables and hence the circular functions need to be defined independently. For circular functions there is the traditional approach of defining them via a circle. Thus if $P$ is a point on the unit circle $x^{2} + y^{2} = 1$ with center at origin $O$ and $A = (1, 0)$ then the coordinates of point $P$ are defined to be $(\cos \theta, \sin \theta)$ where $\theta$ is the length of arc $AP$ (or $\theta$ is twice the area of sector $AOP$, this area approach is simpler).

Another approach is to define $\arctan x$ via integral $\int_{0}^{x}dt/(1 + t^{2})$ and then invert them. It is also possible to define the circular functions by their infinite series representation or infinite product representation, but then these are more popular when we want to examine them as functions of a complex variable.

Once the elementary functions are defined for real variables, it is a trivial matter to define them for complex variable (just as once we define real numbers in terms of rationals, the next step to define complex numbers in terms of real numbers is trivial). Thus for example we can define $$\exp(x + iy) = \exp(x)(\cos y + i \sin y)$$ The link between circular and exponential/logarithmic functions cannot be appreciated unless one defines them as function of a complex variable.

Update: Various approaches to the theory of circular functions (as mentioned above) are now available on blog:


You can proceed through steps 3-5 but reverse steps 1 and 2. One way to do this is to define $\exp$ as the unique solution to $y'=y,y(0)=1$. Proceeding this way takes some work, because you need to prove the Picard-Lindelof theorem to ensure that there is a unique solution in the first place. Still, once you do that, you have $\exp$.

Next, positivity of $\exp$ follows from uniqueness: the solution to $y'=y,y(0)=0$ is $y \equiv 0$ and is unique. Also, the ODE is autonomous. Consequently $\exp$ cannot cross $y=0$. Hence $\exp$ is monotone, so it has an inverse which is defined on the range of $\exp$; call this inverse $\ln$.

The last thing to do is to show that the range of $\exp$ is $(0,\infty)$. First, taking two derivatives gives convexity, which reveals $e^x \geq 1+x$; so the limit at $+\infty$ is $+\infty$. The last thing is to show that the limit at $-\infty$ is $0$; this can be shown by demonstrating the functional equation $\exp(t+s)=\exp(t)\exp(s)$, which follows from the ODE again.

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    $\begingroup$ You don't need the full hammer of Picard-Lindelof to show uniqueness of the solution to $y'=y,\ y(0)=1$. It's like a 5 line proof to show uniqueness in this particular case. I'm not sure how to show existence in an easy way, but beginning calculus students usually won't question a professor who says that something exists. $\endgroup$
    – user137731
    Commented Jan 1, 2016 at 15:14
  • $\begingroup$ @Bye_World: Can you give a hint for a short uniqueness proof? :) Thanks. :) $\endgroup$ Commented Jan 2, 2016 at 11:51
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    $\begingroup$ @H.R.: Assume $y' = y$ on the set of reals and $y(0) = 1$, consider $f(x) = y(x)/e^{x}$, and use the quotient rule and identity theorem to show $f(x) = 1$ for all $x$. $\endgroup$ Commented Jan 2, 2016 at 16:32
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    $\begingroup$ @H.R. The quotient rule applied to the $f$ given by Andrew yields $$f'(x) = \frac{y'(x)e^x-y(x)e^x}{(e^x)^2}$$ Using the characterization of $y$ what can you conclude? $\endgroup$
    – user137731
    Commented Jan 2, 2016 at 16:53
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    $\begingroup$ [email protected] Except you need to already know $e^x$ never vanishes to define $f(x)=y(x)/e^x$, which is one of the things I was trying to prove in the first place. You also need to know that it is equal to its own derivative to do this calculation...but I haven't even defined it yet in this approach. So working around Picard-Lindelof would require a significantly different approach, I think. $\endgroup$
    – Ian
    Commented Jan 3, 2016 at 1:58

In my opinion the most natural and intuitive progression is as follows. I thought this was the standard, but maybe it is not based on the other answers here.

First you have addition. Then multiplication is repeated addition. Then raising a number to a power is repeated multiplication, $$f(b,x) = b^x = \underbrace{b \cdot b \dots b}_{x \text{ times}}$$ In this function of two variables first you can think of $f(b,x)$ as a function of $b$ with $x$ fixed, getting power functions $b \mapsto b^x$. But naturally one would also consider keeping $b$ fixed and considering $f(b,x)$ as a function of $x$, yielding the exponential function (with base $b$), $x \mapsto b^x$.

These can all be motivated based on natural numbers, then extended to integers, then to rationals, and finally extended (or handwaved depending on the level of the audience) to all reals.

Logarithms with respect to a base $b$ is the inverse of the exponential function $b^x$ with that base. That is, $x=\log_b(y)$ is the number such that $b^x=y$, which is unique by monotonicity.

Once exponential functions with a general base are established, one can define the exponential function based on the principle of continuously compounded interest. If you compound interest half as much, but do it twice as much, how does the result change? What about if you compound one tenth as much but do it ten times? One defines the exponential function as the limit where the interest is compounded less and less, but done more and more often: $$\exp(x) := \lim_{n \rightarrow \infty} \left(1 + \frac{x}{n}\right)^{n}.$$ Then after some algebraic manipulations and interchange of limits, one sees that $\exp(x) = e^x$ is indeed an exponential with the particular base $e$ defined as, $$e:=\lim_{n \rightarrow \infty}\left(1 + \frac{1}{n}\right)^n.$$

Likewise, the natural logarithm is defined as the logarithm with respect to base $e$.

At the same time, trigonometric functions $\sin, \cos, \tan$, etc, are defined based on the geometry of right triangles on the unit circle, in the standard way. If you start at the point $(1,0)$ and walk around the unit circle by angle $\theta$, then $\cos(\theta)$ is your $x$ coordinate and $\sin(\theta)$ is your $y$ coordinate - you end out at the point $(\cos(\theta), \sin(\theta))$. Similarly, if you start at the point $(0,1)$ and rotate by angle $\theta$, you are now at the point $(-\sin(\theta),(\cos(\theta)))$.

But rotations are linear - if you rotate two vectors and add them, it is the same as adding them beforehand and then rotating. Thus the action of rotation on the plane is uniquely defined by how it acts on the unit basis vectors, leading to the rotation matrix, $$\text{rotate}_\theta\left(\begin{bmatrix}a \\ b\end{bmatrix}\right)= \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta)\end{bmatrix}\begin{bmatrix}a \\ b\end{bmatrix} = \begin{bmatrix}a \cos(\theta) - b \sin(\theta) \\ a \sin(\theta) + b \cos(\theta)\end{bmatrix}.$$

On the other hand, multiplication by a complex number is also a linear operation based on the distributive law of complex multiplication, and multiplying by a complex number $\cos(\theta) + i \sin(\theta)$ on the unit circle on the complex plane has exactly the same action as rotation by angle $\theta$ (if you identifying points in the plane $\mathbb{R}^2$ with the complex plane $\mathbb{C}$). That is, $$\left(\cos(\theta) + i \sin(\theta)\right)(a + bi) = \left(a \cos(\theta)-b \sin(\theta)\right) + i\left(a \sin(\theta) + b \cos(\theta)\right)$$

From this it is straightforward to deduce the standard rule that multiplication of two complex numbers multiplies their magnitudes and adds their angles.

If we have a complex number that is nearly on the unit circle, then its action is nearly a rotation: $$\left(1 + \frac{i\theta}{n}\right)z \approx \text{rotate}_\frac{\theta}{n}(z).$$

Hence the natural extension of our definition of the exponential function to complex numbers acts as a rotation, $$\exp(i \theta) = \lim_{n \rightarrow \infty}\left(1 + \frac{i \theta}{n}\right)^n$$ since its action on a complex number $z$ is the accumulation of many very small rotations: $$\left(1 + \frac{i \theta}{n}\right)^n z \approx \underbrace{\left(\text{rotate by } \frac{\theta}{n}\right)\left(\text{rotate by } \frac{\theta}{n}\right) \dots \left(\text{rotate by } \frac{\theta}{n}\right)}_{n \text{ times}}z = \left(\text{rotate by } \theta\right) z.$$

Since we already determined that the complex number that rotates by angle $\theta$ is $\cos(\theta) + i \sin(\theta)$, we arrive at Euler's formula connecting complex exponentials and trigonometric functions $$\exp(ix) = \cos(x) + i \sin(x).$$ Everything in the preceeding discussion can be made precise and rigorous by following the same steps but keeping careful of epsilons and limits.

The hyperbolic sine and cosine are just slices of the extended sine and consine functions along the imaginary axis, and are somewhat of a sidenote in this treatment.

In summary,

  • Exponentiation is repeated multiplication.
  • $\exp(x)$ (or $\exp(z)$ for complex numbers) is the function you get by multiplying smaller and smaller fractions of a number more and more times in a continuously compounding limit.
  • There exists a number $e$ such that the exponential function is actually exponential with that base: $\exp(x) = e^x$.
  • The logarithm is the inverse function of the exponential
  • Trigonometric functions like sine, cosine, tangent, etc are defined in the standard geometric way based on rotations in the unit circle.
  • The fact that multiplying by complex numbers scales the lengths and adds the angles is derived from purely geometric/linear-algebraic concepts.
  • The complex exponential and the trigonometric functions are related through Euler's formula, which is a natural consequence of the fact that multiplying by a unit length complex number is a rotation.
  • Hyperbolic sine and consine are slices of sine and cosine along the imaginary axis, and are somewhat of a sidenote in this treatment.

This answer can't do as it relies on the theory of complex functions. Anyway it all boils down to the exponential $e^z.$

You can define it starting from the differential equation $\exp'(z)=\exp(z)$ or from the limit definition $\lim_{n\to\infty}{(1+\dfrac zn)^n}$. And from one of these, derive the Taylor development.

By inversion you get the logarithm $\ln(z)$ and/or the differential equation $\ln'(z)=\dfrac1z$.

At this stage, it is worth to establish the formula for the exponential of a sum, and bridge the gap with ordinary powers.

Then, exploring the pure imaginary arguments of the exponential, you define the trigonometric functions by the Euler formula $e^{is}=:\cos(s)+i\sin(s)$.

You can make a link to geometry by showing that the equation of the unit circle $\cos^2(s)+\sin^2(s)=1$ holds. Then the connection to the angle, or more precisely the curvilinear abscissa along the circle, is obtained by integrating $ds=\sqrt{dx^2+dy^2}$.

From these building blocks, it is an easy matter to define the whole set of direct/inverse trigonometric/hyperbolic functions. As byproducts, you also get the polar forms of the complex exponential and logarithm.


There are certainly many different approaches.

You can start by defining $e^x$ using it's Taylor series.

Then notice that the series still make sense over the complex numbers (i.e. converges everywhere). Then you can define $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ and $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$.

You then define the rest of the trig functions and hyperbolic trig functions in the usual way from $e^x$,$\sin(x)$ and $\cos(x)$ and define $\ln(x)$ as the inverse to $e^x$.

You can also start by defining $e^x$ as the unique function such that $f^\prime(x)=f(x)$, $f(0)=1$ and which is continuous everywhere.

You can define $\sin(x)$ and $\cos(x)$ using the functional equations they satisfy and continuity (you might also need to fix a point though I'm not sure).

$$(f(x))^2+(g(x))^2=1$$ $$f(-x)=-f(x)\wedge g(-x)=g(x)$$ $$f(x+y)=f(x)g(y)+f(y)g(x)$$ $$g(x+y)=g(x)g(y)-f(x)f(y)$$

And go from there using complex numbers to define $e^x$.

You can also define $a^x$ for all real $x$ and $a>0$ by first defining powers of rational exponents in the usual way and then extending to all real numbers density of rationals and continuity. And then pick out $e$ using some limit for example.

You can also define the trig functions using the usual geometrical approach (rigorously) and then extending them in the sensible way to all reals.

And there are many many more ways.

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    $\begingroup$ @H.R. I have the squares on the outside of the parenthesis intentionally. Otherwise you get ambiguity between squaring and functional composition. $\endgroup$
    – DRF
    Commented Dec 4, 2015 at 14:29

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