# Complex numbers; trigonometric properties

I have trouble doing this exercise:

Let $z = cosθ + isinθ$ Expand $(z + z^{−1})^6 and (z-z^{−1})^6$ Hence show that $cos^6θ+sin^6θ=1/8(3cos4θ+5)$

I expanded these two brackets and I got: $2cos6θ+12cos4θ+30cos2θ+20$ for the first one and $2cos6θ-12cos4θ+30cos2θ-20$ for the second one

Is that correct?

I also know that $(z + z^{−1})^6$ is $64cos^6θ$ and $(z-z^{−1})^6$ is $64sin^6θ$

but when I add these two result it doesn't work

Where do I make the mistake?

Thanks for help!

• I reckon that $(z-z^{-1})^6$ is actually $-64\sin^6 \theta$, May 15 '17 at 17:31

$$z-z^{-1}=2i\sin\theta\implies(z-z^{-1})^6=(2\sin\theta)^6i^6=-64\sin^6\theta$$
$$64(\cos^6\theta+\sin^6\theta)=(z+z^{-1})^6-(z-z^{-1})^6=2\left[\binom61\left(z^4+z^{-4}\right)+\binom63\right]$$
• @Markowska, Have you noticed $64$ in the left hand side? May 15 '17 at 17:48