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

Question

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 $$a = 1 + \cos x + \cos2x + \cos3x \quad\text{and}\quad b = \sin x + \sin2x + \sin3x$$

Answer 1

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}

Answer 2

Hint:

$$1+\text{cis}2y=1+\cos2y+i\sin2y=2\cos y(\cos y+i\sin y)$$

$$|1+\text{cis}2y|=|2\cos y|\cdot|\cos y+i\sin y|=2|\cos y|$$

$$|1+\text{cis}2y|^2=4|\cos y|^2=4\cos^2y$$