We present a way to define $\sin$ and $\cos$ which is quite traditional, but show a non-canonical way to "prove" that these definitions are equivalent to the geometrical ones.

First, let's define the derivative of a function $f:\mathbb{R} \rightarrow \mathbb{C}$:

$f'(x)=\Re'(f(x))+i \Im'(f(x))$

Now, extend the definition of exponentiation (read the first post on this blog) to complex numbers:

The series converges for every complex $z$ by the ratio test, and the formula $e^{(z+w)}=e^ze^w$ still holds by the cauchy product formula. Now, let's calculate the derivative of $e^x$ and $e^{ix}$. Note that $x$ is real.

It's common to do this by theorems of power series. We shall not use them. Instead, we use more elementary methods.

For the derivative of $e^x$:

$\displaystyle \lim_{h\rightarrow 0} \frac{e^{x+h}-e^x}{h}=e^{x}\lim_{h\rightarrow 0} \frac{e^h-1}{h}$

Now, to evaluate the last limit (without using theorems of power series), do the following:

Fix an arbitrary $H >0$.

Now, given an $\epsilon >0$, there exists $n \in \mathbb{N}$ such that:

$$\frac{H^{n}}{(n+1)!}+\frac{H^{n+1}}{(n+2)!}+... \leq \epsilon$$

since the series $\displaystyle \sum_{k=0}^{\infty}\frac{H^k}{(k+1)!}$ converges by the ratio test. But note that if you multiply $0<h<H$ this implies :

$$\frac{hH^{n}}{(n+1)!}+\frac{hH^{n+1}}{(n+2)!}+... \leq \epsilon.h$$

Since $h<H$:

$$\frac{h^{n+1}}{(n+1)!}+\frac{h^{n+2}}{(n+2)!}+... \leq \frac{hH^{n}}{(n+1)!}+\frac{hH^{n+1}}{(n+2)!}+... \leq \epsilon.h$$

But then, we have:

$$e^h \leq 1+h+\frac{h^2}{2!}+\frac{h^3}{3!}+...+\frac{h^n}{n!} + \epsilon.h$$

Which gives us:

$$\frac{e^h -1}{h} \leq 1+\frac{h}{2!}+\frac{h^2}{3!}+...+\frac{h^{n-1}}{n!} + \epsilon$$

But $1\leq \frac{e^h -1}{h}$ is obvious from the definition of $e^h$. So, taking limits:

$$1 \leq \displaystyle \lim_{h\rightarrow 0^{+}} \frac{e^h -1}{h} \leq 1+\epsilon$$

But $\epsilon>0$ was arbitrary, which gives:

$$\lim_{h\rightarrow 0^{+}} \frac{e^h -1}{h} =1$$

Now, note that:

$$\displaystyle \lim_{h\rightarrow 0^{-}} \frac{e^h -1}{h} =

\lim_{h\rightarrow 0^{+}} \frac{e^{-h}-1}{-h}= \lim_{h\rightarrow 0^{+}} \frac{\frac{1}{e^h}-1}{-h}=

\lim_{h\rightarrow 0^{+}} \frac{e^h-1}{h}.\frac{1}{e^h}=1$$

Hence, the limit equals $1$, and it is proved that the derivative of $e^x$ is $e^x$. $\blacksquare$

Now, we will calculate the derivative of $e^{ix}$:

$\displaystyle \lim_{h\rightarrow 0} \frac{e^{i(x+h)}-e^{ix}}{h}=e^{ix}\lim_{h\rightarrow 0} \frac{e^{ih}-1}{h}=e^{ix}\lim_{h\rightarrow 0} \frac{e^{ih}-1}{h}$.

But $e^{ih}=1+ih-\frac{h^2}{2!}-i\frac{h^3}{3!}+\frac{h^4}{4!}+...$. Since the series is absolutely convergent, separate the series in two pieces: the part with $i$ and the part without $i$. Similar estimations that were used before now will be able to be used, and will result (since the term in $h$ is $i$):

$$\lim_{h\rightarrow 0} \frac{e^{ih}-1}{h}=i$$

So, the derivative of $e^{ix}$ is $ie^{ix}$.

You may ask at this point: where is $\cos$ and $\sin$?

$\displaystyle \cos(x):=\frac{e^{ix}+e^{-ix}}{2}$

$\displaystyle \sin(x):=\frac{e^{ix}-e^{-ix}}{2i}$

By the definition of $e^z$, $e^{\overline{z}}=\overline{e^z}$. Then, $\cos$ and $\sin$ are real functions. Moreover, it is evident that:

$$e^{ix}=\cos(x)+ i \sin(x)$$

We also have:

$$|e^{ix}|^2=e^{ix}.\overline{e^{ix}}=e^{ix}e^{-ix}=1$$

which implies:

$$|e^{ix}|=1 \Rightarrow \sin^2(x)+\cos^2(x)=1$$

Also, directly from definition:

$$\cos'(x)=-\sin(x), ~~~~~~\sin'(x)=\cos(x)$$

And also directly from definition: $\cos(0)=1$, $\sin(0)=0$

Now, why on earth are those definitions the sine and cosine we know?

We will prove they

(1) $c(0)=1$, $s(0)=0$

(2)$c'(x)=-s(x)$, $s'(x)=c(x)$.

So, $s(x)=\sin(x)$ and $c(x)=\cos(x)$.

This way, since the functions sine and cosine we know geometrically satisfy those properties, they must be the $\sin$ and $\cos$ we just defined.

Define the function $f(x):=(\cos(x)-c(x))^2+(\sin(x)-s(x))^2$. We have:

$$f'(x)=2(\cos(x)-c(x))(-\sin(x)+s(x))+2(\sin(x)-s(x))(\cos(x)-c(x))=0$$

Therefore, $f$ is constant.

But $f(0)=(1-1)^2+(0-0)^2=0$. So $f(x)=0$ for all $x \in \mathbb{R}$.

But this can only be true if $\sin(x)=s(x)$ and $\cos(x)=c(x)$ for all $x \in \mathbb{R}$. $\blacksquare$.

First, let's define the derivative of a function $f:\mathbb{R} \rightarrow \mathbb{C}$:

**Definition:**Given a function $f:\mathbb{R} \rightarrow \mathbb{C}$ given by: $f(x)=\Re(f(x))+i \Im(f(x))$, define:$f'(x)=\Re'(f(x))+i \Im'(f(x))$

**OBS**: Note that theorems like "derivative of sum is sum of derivatives" still hold, as well as the definition of derivative by the limit.**OBS:**Note also that this ISN'T the derivative of a function $f:\mathbb{C} \rightarrow \mathbb{C}$. We are concerned with functions with real domain.Now, extend the definition of exponentiation (read the first post on this blog) to complex numbers:

**Definition:**$\displaystyle e^z:=\sum_{n=0}^{\infty}\frac{z^n}{n!}$The series converges for every complex $z$ by the ratio test, and the formula $e^{(z+w)}=e^ze^w$ still holds by the cauchy product formula. Now, let's calculate the derivative of $e^x$ and $e^{ix}$. Note that $x$ is real.

It's common to do this by theorems of power series. We shall not use them. Instead, we use more elementary methods.

For the derivative of $e^x$:

$\displaystyle \lim_{h\rightarrow 0} \frac{e^{x+h}-e^x}{h}=e^{x}\lim_{h\rightarrow 0} \frac{e^h-1}{h}$

Now, to evaluate the last limit (without using theorems of power series), do the following:

Fix an arbitrary $H >0$.

Now, given an $\epsilon >0$, there exists $n \in \mathbb{N}$ such that:

$$\frac{H^{n}}{(n+1)!}+\frac{H^{n+1}}{(n+2)!}+... \leq \epsilon$$

since the series $\displaystyle \sum_{k=0}^{\infty}\frac{H^k}{(k+1)!}$ converges by the ratio test. But note that if you multiply $0<h<H$ this implies :

$$\frac{hH^{n}}{(n+1)!}+\frac{hH^{n+1}}{(n+2)!}+... \leq \epsilon.h$$

Since $h<H$:

$$\frac{h^{n+1}}{(n+1)!}+\frac{h^{n+2}}{(n+2)!}+... \leq \frac{hH^{n}}{(n+1)!}+\frac{hH^{n+1}}{(n+2)!}+... \leq \epsilon.h$$

But then, we have:

$$e^h \leq 1+h+\frac{h^2}{2!}+\frac{h^3}{3!}+...+\frac{h^n}{n!} + \epsilon.h$$

Which gives us:

$$\frac{e^h -1}{h} \leq 1+\frac{h}{2!}+\frac{h^2}{3!}+...+\frac{h^{n-1}}{n!} + \epsilon$$

But $1\leq \frac{e^h -1}{h}$ is obvious from the definition of $e^h$. So, taking limits:

$$1 \leq \displaystyle \lim_{h\rightarrow 0^{+}} \frac{e^h -1}{h} \leq 1+\epsilon$$

But $\epsilon>0$ was arbitrary, which gives:

$$\lim_{h\rightarrow 0^{+}} \frac{e^h -1}{h} =1$$

Now, note that:

$$\displaystyle \lim_{h\rightarrow 0^{-}} \frac{e^h -1}{h} =

\lim_{h\rightarrow 0^{+}} \frac{e^{-h}-1}{-h}= \lim_{h\rightarrow 0^{+}} \frac{\frac{1}{e^h}-1}{-h}=

\lim_{h\rightarrow 0^{+}} \frac{e^h-1}{h}.\frac{1}{e^h}=1$$

Hence, the limit equals $1$, and it is proved that the derivative of $e^x$ is $e^x$. $\blacksquare$

Now, we will calculate the derivative of $e^{ix}$:

$\displaystyle \lim_{h\rightarrow 0} \frac{e^{i(x+h)}-e^{ix}}{h}=e^{ix}\lim_{h\rightarrow 0} \frac{e^{ih}-1}{h}=e^{ix}\lim_{h\rightarrow 0} \frac{e^{ih}-1}{h}$.

But $e^{ih}=1+ih-\frac{h^2}{2!}-i\frac{h^3}{3!}+\frac{h^4}{4!}+...$. Since the series is absolutely convergent, separate the series in two pieces: the part with $i$ and the part without $i$. Similar estimations that were used before now will be able to be used, and will result (since the term in $h$ is $i$):

$$\lim_{h\rightarrow 0} \frac{e^{ih}-1}{h}=i$$

So, the derivative of $e^{ix}$ is $ie^{ix}$.

You may ask at this point: where is $\cos$ and $\sin$?

**Definition:**$\displaystyle \cos(x):=\frac{e^{ix}+e^{-ix}}{2}$

$\displaystyle \sin(x):=\frac{e^{ix}-e^{-ix}}{2i}$

By the definition of $e^z$, $e^{\overline{z}}=\overline{e^z}$. Then, $\cos$ and $\sin$ are real functions. Moreover, it is evident that:

$$e^{ix}=\cos(x)+ i \sin(x)$$

We also have:

$$|e^{ix}|^2=e^{ix}.\overline{e^{ix}}=e^{ix}e^{-ix}=1$$

which implies:

$$|e^{ix}|=1 \Rightarrow \sin^2(x)+\cos^2(x)=1$$

Also, directly from definition:

$$\cos'(x)=-\sin(x), ~~~~~~\sin'(x)=\cos(x)$$

And also directly from definition: $\cos(0)=1$, $\sin(0)=0$

Now, why on earth are those definitions the sine and cosine we know?

We will prove they

**must**be. How?**Proposition:**Let $c:\mathbb{R} \rightarrow \mathbb{R}$ and $s: \mathbb{R} \rightarrow \mathbb{R}$ such that:(1) $c(0)=1$, $s(0)=0$

(2)$c'(x)=-s(x)$, $s'(x)=c(x)$.

So, $s(x)=\sin(x)$ and $c(x)=\cos(x)$.

This way, since the functions sine and cosine we know geometrically satisfy those properties, they must be the $\sin$ and $\cos$ we just defined.

*Proof:*Suppose we have functions $c, s$ satisfying those properties.Define the function $f(x):=(\cos(x)-c(x))^2+(\sin(x)-s(x))^2$. We have:

$$f'(x)=2(\cos(x)-c(x))(-\sin(x)+s(x))+2(\sin(x)-s(x))(\cos(x)-c(x))=0$$

Therefore, $f$ is constant.

But $f(0)=(1-1)^2+(0-0)^2=0$. So $f(x)=0$ for all $x \in \mathbb{R}$.

But this can only be true if $\sin(x)=s(x)$ and $\cos(x)=c(x)$ for all $x \in \mathbb{R}$. $\blacksquare$.