Trigonometry or Real or a Complex Problem

Geometry Level 4

If $2 \cos x = z + \dfrac{1}{z}$ and $z^{12} + \dfrac{1}{z^{12}} = 2$ , which of the following options is true?

\cos 6x\in\{1,-1\}, \cos 9x\in\{1,-1\}

\cos 3x\in\{-1,1\}, \cos 9x\in \{-1,1\}

\sin 2x\in\left\{0,\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}\right\}, \cos 9x\in\{1,-1\}

\cos 6x\in\{-1,1\}, \sin 2x\in\left\{0,\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}\right\}

1 solution

Andreas Wendler
Jun 9, 2016

We write: $z=r\cdot e^{i\phi}$ From the second equation we get: $(r^{12}+r^{-12})cos(12\phi)=2$ as well as $(r^{12}-r^{-12})sin(12\phi)=0$ Solution is r=1 and $\phi=\frac{k\pi}{6}$ whereat k is an integer.

The first equation now gives us $x=\frac{k\pi}{6}$ . From the given choices only the last one is valid!

Trigonometry or Real or a Complex Problem

1 solution

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