Connecting trigonometry with exponential functions


Nobel Prize-winning physicist Richard Feynman called Euler's equation,

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

"the most remarkable formula in mathematics." I think Feynman was right. There are few equations more useful in higher math. The ability to inter-convert between trigonometric and exponential representations of the same function is immensely valuable.

Euler's formula was discovered by Swiss mathematician Leonhard Euler (1707-1783) [pronounced oy'-ler]. If you get a chance, Euler's life in mathematics and science is worth reading about. Few have made the range of contributions he did. In this section we'll prove Euler's formula and use it to link unit-circle trigonometry with hyperbolic trig. (hyperbolic functions).

Euler's formula

$$e^{±ikx} = cos(kx) ± i\, sin(kx),$$


where k is a constant.


A proof (or derivation) of Euler's equation


There are a few ways to arrive at Euler's equation, but we'll do it by finding the MacLaurin series (Taylor series centered at x = 0) of   $f(x) = e^{ix}.$ First we'll need to calculate a few derivatives and find their values at x = 0. Here they are:

$f(x) = e^{ix}$ $f(0) = 1$
$f'(x) = ie^{ix}$ $f'(0) = i$
$f''(x) = -e^{ix}$ $f''(0) = -1$
$f'''(x) = -i e^{ix}$ $f'''(0) = -i$
$f^{(4)}(x) = e^{ix}$ $f^{(4)}(0) = 1$
$f^{(5)}(x) = ie^{ix}$ $f^{(5)}(0) = i$

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Now recall that the MacLaurin series representation of a function is

$$f(x) \approx \frac{f(0)x^0}{0!} + \frac{f'(0)x^1}{1!} + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \dots$$

Plugging in our derivatives gives

$$e^{ix} \approx \frac{1 x^0}{0!} + \frac{i x^1}{1!} + \frac{-1 x^2}{2!} + \frac{-i x^3}{3!} + \frac{1 x^4}{4!} + \frac{i x^5}{5!} + \frac{-i x^6}{6!} + \dots$$

A little bit of clean-up gives us a neater expression:

$$e^{ix} \approx 1 + \frac{i x}{1!} - \frac{x^2}{2!} - \frac{i x^3}{3!} + \frac{x^4}{4!} + \frac{i x^5}{5!} - \frac{i x^6}{6!} - \frac{i x^7}{7!} + \dots $$

Now if we separate terms containing and not containing i, we get two series:

$$e^{ix} \approx \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{i x^6}{6!} + \dots \right) + \left( \frac{i x}{1!} - \frac{i x^3}{3!} + \frac{i x^5}{5!} - \frac{i x^7}{7!} + \dots \right)$$

Now the first parentheses, containing the even-power terms, is the MacLaurin series representation of cos(x). The second is i times the MacLaurin series representation of sin(x):

$$e^{ix} \approx \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{i x^6}{6!} + \dots \right) + i\left( \frac{x}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots \right),$$

which gives us Euler's equation,

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


Exponential expressions for $sin(x)$ and $cos(x)$


To obtain the exponential formula for cos(x), add the Euler equations for $e^{ix}$ and $e^{-ix}.$

$$ \begin{align} & \phantom{00000} e^{ix} = cos(x) + i\, sin(x) \\[5pt] &\underline{+ \phantom{000} e^{-ix} = cos(x) - i\, sin(x)} \\[5pt] & e^{ix} + e^{-ix} = 2 \, cos(x) \end{align}$$

The sine terms cancel, and if we divide by 2, we get

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

The formula for sin(x) is found first by rearranging both Euler equations to solve for cos(x),

$$ \begin{align} cos(x) &= e^{ix} - i \, sin(x) \\[5pt] cos(x) &= e^{-ix} + i \, sin(x) \end{align}$$

Then we eliminate cos(x) between them using the transitive property (if a = c and b = c, then a = b). After that, just divide by 2i to get sin(x).

$$ \begin{align} e^{ix} - i \, sin(x) &= e^{-ix} + i \, sin(x)\\[5pt] e^{ix} - e^{-ix} &= 2i \, sin(x) \\[10pt] sin(x) &= \frac{e^{ix} - e^{-ix}}{2 i} \end{align}$$

Expressions for sin(x) and cos(x)

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

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


The Euler identity


If we use Euler's equation to calculate $e^{i\pi},$ we get

$$e^{i\pi} = cos(\pi) + i \, sin(\pi) = -1$$

$$e^{i\pi} = -1$$

This interesting identity is really pretty amazing. It's a combination of the transcendental number π, the imaginary number i and the base of all continuously-growing exponential functions, e – an eclectic mix, for sure.

Interesting as it is, this expression isn't usually as useful as Euler's equation, but it's worth knowing, if only to amaze your friends.


Relationship to hyperbolic trig. functions


We can use the exponential expressions for sine and cosine to relate those functions to the hyperbolic sine and cosine, sinh(x) and cosh(x).

Consider $sin(ix)$ in its exponential representation:

$$ \begin{align} sin(x) &= \frac{e^{ix} - e^{-ix}}{2 i} \\[5pt] sin(ix) &= \frac{e^{iix} - e^{-iix}}{2i} \\[5pt] &= \frac{e^{-x} - e^x}{2i} \\[5pt] &= \left( \frac{-i}{-i}\right) \frac{e^{-x} - e^x}{2i} \\[5pt] &= i \, sinh(x) \end{align}$$

We can do a similar procedure with $cos(ix).$

$$ \begin{align} cos(x) &= \frac{e^{ix} + e^{-ix}}{2} \\[5pt] sin(ix) &= \frac{e^{iix} + e^{-iix}}{2} \\[5pt] &= \frac{e^{-x} + e^x}{2} \\[5pt] &= cosh(x) \end{align}$$

Unit-circle ↔ hyperbolic

$$sin(ix) = i sinh(x)$$

$$cos(ix) = cosh(x)$$


Practice problems

1.

The trigonometric sum identities for sin(a + b) and cos(a + b) are difficult to derive geometrically, but they are fairly straightforward if you use Euler's equation for sin(x) and cos(x). Derive   $sin(a + b) = cos(a) sin(b) + sin(a) cos(b)$   using Euler's formula.

2.

Find the indefinite integral   $\int cos(ax) \, cos(bx) \, dx,$   where a and b are constants.

3.

Integrate   $\int sin^2(x) \, dx.$

4.

Integrate   $\int sin(2x) sin(4x) \, dx.$

5.

Prove DeMoivre's formula,   $(cos(\theta) + i\, sin(\theta))^n = cos(n\theta) + i\, sin(n\theta)$

6.

Find the value of $i^{-i}$ by writing the imaginary number $i$ as a complex exponential.

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