If $(1 + \operatorname{cis}x)(1 + \operatorname{cis} 2x) = a + bi$, prove that $a^2 + b^2 = (4\cos x \cos\frac x2)^2$ Could anyone help me with this? I'm stuck.

If $$(1 + \operatorname{cis}x)(1 + \operatorname{cis} 2x) = a + bi$$ prove that $$a^2 + b^2 = \left(4\cos x \cos\frac x2\right)^2$$

For reference, $\operatorname{cis}x = \cos x + i\sin x$.
I found that 
$$a = 1 + \cos x + \cos2x + \cos3x \quad\text{and}\quad 
b = \sin x + \sin2x + \sin3x$$

Note
\begin{align}
a^2+b^2 &= |a + bi|^2 \\
&= |(1 + \operatorname{cis}x)(1 + \operatorname{cis} 2x) |^2\\
 &= |1 + e^{i x}|^2|1 + e^{i 2x}|^2\\
 &= |e^{-\frac x2}+ e^{-\frac x2}|^2 |e^{-i x}+ e^{ ix}|^2\\
 &= |2\cos\frac x2|^2 |2\cos x|^2\\
&= (4\cos x \cos\frac x2)^2
\end{align}
