Show that $e^{i \theta} + e^{-i \theta} = 2 cos \theta$. Hence show that $\mathrm{cos}^3 \theta = \frac{1}{4} (\mathrm{cos}3 \theta + 3 \mathrm{cos} \theta)$

For the first part, we can simply apply Euler’s Formula, that is $e^{i \theta} = \mathrm{cos} \theta + i \mathrm{sin} \theta$

$e^{i \theta} + e^{-i \theta}$
$= \mathrm{cos} \theta + i \mathrm{sin} \theta + \mathrm{cos} (- \theta) + i \mathrm{sin} (- \theta)$
$= \mathrm{cos} \theta + i \mathrm{sin} \theta + \mathrm{cos} \theta - i \mathrm{sin} \theta$
$= 2 \mathrm{cos} \theta$

The next part is a little more tricky, and since its hence, we will use what we solved previously to help us.

$e^{i \theta} + e^{-i \theta} = 2 cos \theta$
$\Rightarrow cos \theta = \frac{1}{2}(e^{i \theta} + e^{-i \theta})$

$\mathrm{cos}^3 \theta$

$= (\mathrm{cos})^3$

$= [\frac{1}{2}(e^{i \theta} + e^{-i \theta})]^3$

$= \frac{1}{8}(e^{i \theta} + e^{-i \theta})^3$

$= \frac{1}{8}(e^{i 3\theta} + 3 e^{i 2\theta}e^{-i \theta} + 3 e^{i \theta}e^{-i 2\theta} + e^{-i 3\theta})$

$= \frac{1}{8}(e^{i 3\theta} + 3 e^{i \theta} + 3 e^{-i \theta} + e^{-i 3\theta})$

$= \frac{1}{8}\{\mathrm{cos} 3\theta + i \mathrm{sin} 3\theta + 3(\mathrm{cos} \theta + i \mathrm{sin} \theta) + 3[\mathrm{cos} (-\theta) + i \mathrm{sin} (-\theta)] + \mathrm{cos} (-3\theta) + i \mathrm{sin} (-3\theta) \}$

$= \frac{1}{8}(\mathrm{cos} 3\theta + i \mathrm{sin} 3\theta + 3\mathrm{cos} \theta + 3i \mathrm{sin} \theta + 3\mathrm{cos} \theta - 3i \mathrm{sin} \theta + \mathrm{cos} 3\theta - i \mathrm{sin} 3\theta)$

$= \frac{1}{8}(\mathrm{cos} 3\theta + i \mathrm{sin} 3\theta + 3\mathrm{cos} \theta + 3i \mathrm{sin} \theta + 3\mathrm{cos} \theta - 3i \mathrm{sin} \theta + \mathrm{cos} 3\theta - i \mathrm{sin} 3\theta)$

$= \frac{1}{8}(\mathrm{cos} 3\theta + 3\mathrm{cos} \theta + 3\mathrm{cos} \theta + \mathrm{cos} 3\theta)$

$= \frac{1}{8}(2\mathrm{cos} 3\theta + 6\mathrm{cos} \theta )$

$= \frac{1}{4}(\mathrm{cos} 3\theta + 3\mathrm{cos} \theta )$

Just for fun…

$2e^{\frac{5\pi}{6}i} + \frac{1}{2e^{\frac{5\pi}{6}i}}$
$= 2e^{\frac{5\pi}{6}i} + \frac{1}{2}2e^{\frac{-5\pi}{6}i}$
$= 2 [\mathrm{cos}(\frac{5\pi}{6}) + i \mathrm{sin}(\frac{5\pi}{6})] + \frac{1}{2 } [\mathrm{cos}(\frac{-5\pi}{6}) + i \mathrm{sin}(\frac{-5\pi}{6})]$
$= 2 (\frac{-\sqrt{3}}{2} + i \frac{1}{2}) + \frac{1}{2} (\frac{-\sqrt{3}}{2} - i \frac{1}{2})$
$= -\sqrt{3} + i + \frac{-\sqrt{3}}{4} - i\frac{1}{4}$
$= -\frac{5\sqrt{3}}{4} + \frac{3}{4}i$

$e^{x+yi} = (1+i)^6 = [\sqrt{2}e^{i\frac{\pi}{4}}]^6$
$e^{x+yi} = 8e^{i\frac{3\pi}{2}}$
$e^x \times e^{yi} = 8 e^{i\frac{3\pi}{2}}$
$\Rightarrow e^x = 8 \Rightarrow x = \mathrm{ln}8 = 3 \mathrm{ln}2$
$\Rightarrow y = \frac{3\pi}{2} - 2\pi = \frac{-\pi}{2}$

Complex Number Problem #1

### One Comment

• […] for students stuck, consider checking this link here for (a) and this link here for (b). These links hopefully enlightens […]