Connecting trigonometry with exponential functions

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

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

"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.

A proof (or derivation) of Euler's equation

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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:

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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 $\text{sin}(x)$ and $\text{cos}(x)$

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

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

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

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

The formula for $\text{sin}(x)$ is found first by rearranging both Euler equations to solve for $\text{cos}(x)$,

The Euler identity

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

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

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

Relationship to hyperbolic trig. functions

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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:

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

We can do a similar procedure with $\text{cos}(ix).$

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 $\text{sin}(x)$ and $\text{cos}(x)$. Derive   $\text{sin}(a + b) = \text{cos}(a) \text{sin}(b) + \text{sin}(a) \text{cos}(b)$   using Euler's formula.

   Solution

   The idea here is to derive both the sine and cosine sum formulae at the same time. We start by writing the exponential expression:

   $$e^{i(a + b)} = cos(a + b) + i \, sin(a + b)$$

   Now   $e^{i(a + b)} = e^{ia} e^{bi}.$   We can apply that same multiplicative separation (we're taking advantage of the unique algebra of exponentials) to the cosine and sine terms:

   $$ \begin{align} e^{ia} e^{bi} &= [cos(a) + i \, sin(a)][cos(b) + i \, sin(b)] \\[5pt] &= \text{cos}(a) \text{cos}(b) + i\, \text{cos}(a) \text{sin}(b) + i\, \text{sin}(a) \text{cos}(b) - \text{sin}(a) \text{sin}(b) \\[5pt] &= \text{cos}(a) \text{cos}(b) - \text{sin}(a) \text{sin}(b) + i[\text{cos}(a) \text{sin}(b) + \text{sin}(a) \text{cos}(b)] \end{align}$$

   Now the real term corresponds with   $\text{cos}(a + b)$   and the imaginary term corresponds to   $\text{sin}(a + b).$   So we have:

   $$ \begin{align} \text{cos}(a + b) = \text{cos}(a) \text{cos}(b) - \text{sin}(a) \text{sin}(b) \\[8pt] \text{sin}(a + b) = \text{cos}(a) \text{sin}(b) + \text{sin}(a) \text{cos}(b) \end{align}$$

   This is a much easier derivation than the geometric one.




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

   Solution

   $$\int \, cos(ax) cos(bx) \, dx$$

   First convert those cosine functions to exponential form. Then multiply the binomials:

   $$ \begin{align} &= \int \left( \frac{e^{ix} + e^{ix}}{2} \right)\left( \frac{e^{ix} + e^{ix}}{2} \right) \, dx \\[5pt] &= \frac{1}{4} \int \left[ e^{i(a + b)x} + e^{i(a - b)x} + e^{i(b - a)x} + e^{-(a + b)x} \right] \, dx \\[5pt] &= \frac{1}{4} \int \left[ e^{i(a + b)x} + e^{i(a - b)x} + e^{-i(a - b)x} + e^{-(a + b)x} \right] \, dx \\[5pt] &= \frac{1}{2} \int \left( \frac{e^{i(a + b)x} + e^{-i(a + b)x}}{2} \right) + \left( \frac{e^{i(a - b)x} + e^{-(a - b)x}}{2} \right) \, dx \\[5pt] &= \frac{1}{2} \int \left( \text{cos}[(a + b)x] + \text{cos}[(a - b)x]\right) \, dx \\[5pt] &= \frac{1}{2} \left[ \frac{1}{a + b}\text{sin}[(a + b)x] + \frac{1}{a - b}\text{sin}[(a - b)x] \right] \\[5pt] &= \frac{\text{sin}[(a + b)x]}{2(a + b)} + \frac{\text{sin}[(a - b)x]}{2(a - b)} + C \end{align}$$

   There are many more ways to manipulate and express this result, but this is a moderate stopping point, and it's the way this integral is usually expressed in integral tables.




3. Integrate   $\int \text{sin}^2(x) \, dx.$

   Solution

   $$\int \, \text{sin}^2(x) \, dx$$

   Express \text{sin}(x) in exponential form, then simplify the integrand, re-convert to a trig. function and integrate:

   $$ \begin{align} &= \int \left( \frac{e^{ix} - e^{-ix}}{2i} \right)^2 \, dx \\[5pt] &= \int \frac{e^{i2x} - 1 - 1 + e^{i2x}}{-4} \, dx \\[5pt] &= -\frac{1}{2} \int \frac{e^{i2x} + e^{-i2x}}{2} + \int \frac{1}{2} \, dx \\[5pt] &= -\frac{1}{2} \int \text{cos}(2x) \, dx + \int \frac{1}{2} \, dx \\[5pt] &= \frac{x}{2} - \frac{1}{4} \text{sin}(2x) + C \end{align}$$

   It's arguable whether using the exponential route makes this integral simpler. Substitution for $\text{sin}^2(x)$ at the beginning using the power-reduction identity, $\text{sin}^2(x) = (1 - \text{cos}(2x))/2,$ makes this integral pretty simple, too. What's great is to have a wide variety of integration tools in your kit. Always remember, though, that not all integrals can be done analytically.




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

   Solution

   $$ \begin{align} \int &\text{sin}(2x) \text{sin}(4x) \, dx \\[5pt] &= \int \left( \frac{e^{i2x} - e^{-i2x}}{2i} \right)\left( \frac{e^{i4x} - e^{-i4x}}{2i} \right)\, dx \\[5pt] &= \int \frac{e^{i6x} - e^{-i2x} - e^{i2x} + e^{-i6x}}{-4}\, dx \\[5pt] &= -\frac{1}{2} \int \left( \frac{e^{i6x} + e^{-i6x}}{2} - \frac{e^{i2x} + e^{-i2x}}{2}\right) \, dx \\[5pt] &= -\frac{1}{2} \int \text{cos}(6x) \, \text{cos}(2x) \, dx \\[5pt] &= -\frac{1}{12} \text{sin}(6x) - \frac{1}{4} \text{sin}(2x) + C \end{align}$$




5. Prove DeMoivre's formula,   $(\text{cos}(\theta) + i\, \text{sin}(\theta))^n = \text{cos}(n\theta) + i\, \text{sin}(n\theta)$

   Solution

   Substitute the exponential expressions into the sine and cosine terms in $(cos(x) + i\, sin(x))^n.$ (I'm switching to x because it's easier to type than "\theta."

   $$ \begin{align} &= \left[\frac{e^{ix} + e^{-ix}}{2} + i \, \left( \frac{e^{ix} - e^{-ix}}{2i} \right)\right]^n\\[5pt] &= \left( \frac{e^{ix}}{2} + \frac{e^{-ix}}{2} + \frac{e^{ix}}{2} - \frac{e^{-ix}}{2} \right)^n \\[5pt] &= \left( e^{ix} \right)^n = e^{nix} \\[5pt] &= \text{cos}(bx) + i\, \text{sin}(bx) \end{align}$$

   DeMoivre's formula is really handy. When we need a large poer of a complex vector, we need only multiply the argument of the sine and cosine by the power.




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

   Solution

   

   $$i^{-i} = \; ?$$

   Use the complex-plane graph (Re = real, Im = imaginary) to re-write the complex vector $z = 0 + i$ with sines and cosines:

   $$z = \text{cos}\left( \frac{\pi}{2} \right) + i \, \text{sin} \left( \frac{\pi}{2} \right) = e^{i \frac{\pi}{2}}$$

   Notice that $\text{cos}(\pi/2) = 0$ and $\text{sin}(\pi/2) = 1,$ so this expression captures our complex vector. Now raise the exponential form to the power of -i

   $$\left(e^{i\frac{\pi}{2}} \right)^{-i} = e^{\frac{\pi}{2}} = 4.81$$

   Doesn't it seem strange that such an expression would have a numerical value?