Expand using binomial theorem

Question

Yesterday, my maths lecturer did this:

$[(\cos x+i\sin x)+(\cos x-i\sin x)]^4$

Expanded to:

$8\cos2x+2\cos4x+6$

I've tried using binomial theorem to get to this answer but I can never seem to get there. How can I do it?

When I do it I get to:

$2\cos4x+4(\cos3x+i\sin3x)(\cos x-i\sin x)+6(\cos^22x+\sin^2(2x))+4(\cos^2x-i^2\sin^2x)$

Answer 1

If we combine like terms, we get $(2\cos x)^4$. The Double-angle formula for $\cos$ is: $$\cos 2x=2\cos^2x-1$$ We start with $(2\cos x)^4=16\cos^4x$ and get $4(2\cos^2x)^2=4(\cos2x+1)^2=4\cos^22x+8\cos 2x+4,$ and the first term gives $2(2\cos^22x)=2\cos4x+2$, for a grand total of $2\cos 4x+8\cos 2x+6$.

From where you left off, it is possible to use $\cos^2 a+\sin^2 a=1$ to reduce a good portion of your binomial results to a simple value of $10$, after which you can focus on the not-yet-distributed portion of your results.

Answer 2

This is

$$(2\cos x)^4=(e^{ix}+e^{-ix})^4$$ which expands as

$$\color{red}1e^{i4x}+\color{green}4e^{i2x}+\color{blue}6+\color{green}4e^{-i2x}+\color{red}1e^{-i4x}.$$ (Notice the binomial weights.)

Regrouping and using the complex definition of the cosine,

$$2\cdot\color{red}1\cos 4x+2\cdot\color{green}4\cos2x+\color{blue}6.$$

The generalization to all even $n$ should be obvious.

Answer 3

Following on from what you've already done:

Using Werner's formulas (see http://mathworld.wolfram.com/WernerFormulas.html) $$\begin{aligned}\cos(3x)\cos x&=\frac{1}{2}(\cos(2x)+\cos(4x))\\ \sin(3x)\sin x&=\frac{1}{2}(\cos(2x)-\cos(4x))\\ \cos(3x)\sin x&=\frac{1}{2}(\sin(4x)-\sin(2x))\\ \sin(3x)\cos x&=\frac{1}{2}(\sin(4x)+\sin(2x))\\ \end{aligned}$$

you can show

$$(\cos(3x)+i\sin(3x))(\cos x+i\sin x) = \cos(2x)+i\sin(2x)$$

This will also work, but involves lots of unnecessary calculation.