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:

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

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

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

This is a much easier derivation than the geometric one.

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

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

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

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.

$$\int \, sin^2(x) \, dx$$

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

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

It's arguable whether using the exponential route makes this integral simpler. Substitution for $sin^2(x)$ at the beginning using the power-reduction identity, $sin^2(x) = (1 - cos(2x))/2,$ 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.

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

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."

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

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.

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

Notice that $cos(\pi/2) = 0$ and $sin(\pi/2) = 1,$ 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?