Simplifying trigonometric products into trigonometric sums

It can be hard to differentiate products of trigonometric functions like this one: \[ \cos{y}\sin{y} \] It is easier when the functions are separated like this: \[ \cos{y}+\sin{y} \] for example. Therefore I will here describe how to transform equations of the first form into equations of the second form using the Euler's formula.

Euler's formula lets us transform the $\sin$ and $\cos$ parts into this: \[ \begin{aligned} \DeclareMathOperator{\e}{e} \cos(y) &= \frac{\e^{iy} + \e^{-iy}}{2} \\ \sin(y) &= \frac{\e^{iy} - \e^{-iy}}{2i} \end{aligned} \]

By doing this we can make use of the rules for products of $\e$, which are preserved for complex numbers: \[ \e^{z_1}\;\e^{z_2} = \e^{z_1+z_2} \] and magically doing this will end up with some parts that can be put back on the regular trigonometric form be reversing Euler's formula.

Worked example

I will now rewrite trigonometric products like this: \[ \sin{3x}\cos{5x} = \frac{1}{2} (\sin{8x} - \sin{2x}). \]

First, I will rewrite the left side using Euler's formula: \[ \left( \frac{\e^{3xi} - \e^{-3xi}}{2i} \right) \left( \frac{\e^{5xi} + \e^{-5xi}}{2} \right). \]

I pull the denominator out in front, and use the product rules of $\e$: \[ \begin{aligned} &\frac{1}{4i} (\e^{3xi + 5xi} + \e^{-3xi + 5xi} - \e^{-3xi + 5xi} - \e{-3xi - 5xi}) \\ = &\frac{1}{4i}(\e^{8xi} + \e^{-2xi} - \e{2xi} - \e^{-8xi}) \end{aligned} \]

I break it up in two parts: \[ \frac{1}{4i} (\e^{8xi} - \e^{-8xi}) + \frac{1}{4i} (\e^{-2xi} - \e^{2xi}) \]

Now what we are trying to do is to get it on the form defined by Euler's formula. Because there is a minus between the $\e$-s in both the first and the second part, we assume that we want to go back to sinus in both cases. Sinus was defined as: \[ \sin(y) = \frac{\e^{iy} - \e^{-iy}}{2i} \] and therefore part of the denominator of $\frac{1}{4i}$, is “pulled” into the equation like this: \[ \frac{1}{2} \left(\frac{\e^{8xi} - \e^{-8xi}}{2i}\right) + \frac{1}{2} \left(\frac{\e^{-2xi} - \e^{2xi}}{2i}\right) \]

Now the “equation” is on the Euler's form, and can be put back as an ordinary trigonometric function. Taking the signing into condition, it ends up like this: \[ \frac{1}{2} (\sin{8x} - \sin{2x}) \] and thereby the first statement is proved.