Complex Number Problem #2

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.

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

$\mathrm{cos}^3 \theta$

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

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

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

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

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

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

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

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

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

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

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

About

KS Teng

KS has been teaching H2/H1 Mathematics and IB mathematics for the more than 10 years. Having taught students from all various Junior Colleges, KS adapts to students' abilities and help them better understand the topics. As someone who loves to teach mathematics and sees it as a truly useful tool in life, KS seeks to enable students to appreciate math. Therefore, his tuition mission is to motivate and cultivate students to be independent and confident thinkers.

Comments

pingbacks / trackbacks

Random Questions from 2017 Prelims #1 – The Culture SG
[…] for students stuck, consider checking this link here for (a) and this link here for (b). These links hopefully enlightens […]
August 20th, 2017 12:00 AM

Cart

Leave a Reply to Random Questions from 2017 Prelims #1 – The Culture SG Cancel reply