Complex q (1 Viewer)

=)(= · Dec 13, 2022

How would you do 10c
I get up to z^6 -15z^4+15z^2-1

5uckerberg · Dec 14, 2022

$\left(\cos{\theta}+i\sin{\theta}\right)^{6}=\cos^{6}\theta+6i\cos^{5}\theta\sin{\theta}-15\cos^{4}\theta\sin^{2}\theta-20i\cos^{3}\theta\sin^{3}\theta+15\cos^{2}\theta\sin^{4}\theta+6i\cos{\theta}\sin^{5}\theta-\sin^{6}\theta$
$\cos{6\theta}=\cos^{6}\theta-15\cos^{4}\theta\sin^{2}\theta+15\cos^{2}\theta\sin^{4}\theta-\sin^{6}\theta$
$\sin{6\theta}=6\cos^{5}\theta\sin{\theta}-20\cos^{3}\theta\sin^{3}\theta+6\cos{\theta}\sin^{5}\theta$
$\cot{6\theta}=\frac{\cos{6\theta}}{\sin{6\theta}}=\frac{\cos^{6}\theta-15\cos^{4}\theta\sin^{2}\theta+15\cos^{2}\theta\sin^{4}\theta-\sin^{6}\theta}{6\cos^{5}\theta\sin{\theta}-20\cos^{3}\theta\sin^{3}\theta+6\cos{\theta}\sin^{5}\theta}$
Here divide by $\sin^{6}\theta$
we will have $\cot{6\theta}=\frac{\cot^{6}\theta-15\cot^{4}\theta+15\cot^{2}\theta-1}{6\cot^{5}\theta-20\cot^{3}\theta+6\cot{\theta}}$

Now onto part iii you said this I reckon one did $\cot{6\theta}=0$
and then had to say for $\cot{6\theta}=0\;we\;need\;to\;have\;\cot^{6}\theta-15\cot^{4}\theta+15\cot^{2}\theta-1=0$ .
I believe the next step you did was $Let\;z\;=\;\cot{\theta}$
This gives us $z^{6}-15z^{4}+15z^{2}-1$
Factorize the equation $\left(z^{2}-1\right)\left(z^{4}-14z^{2}+1\right)$
Recall when we said $\cot{6\theta}=0\;is\;just\;\frac{1}{\tan{6\theta}}=0\;implying\;\tan{6\theta}=undefined$
The range of $\cot{\theta}\;is\;0\leq{\theta}\leq{\pi}\;so\;\cot{6\theta}\;is\;0\leq{\theta}\leq{6\pi}$
Going back to when we said $\tan{6\theta}=undefined$ then $6\theta=\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{9\pi}{2}, \frac{11\pi}{2}$ , so theoretically $\theta=\frac{\pi}{12}, \frac{3\pi}{12}, \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{9\pi}{12}, \frac{11\pi}{12}$
Note that $\theta=\frac{3\pi}{12}, \frac{9\pi}{12}$ have already being taken by $\left(z^{2}-1\right)$ , so in this equation we are focusing on $\left(z^{4}-14z^{2}+1\right)$ . Hence, $\cot{\frac{\pi}{12}}$ is a root of the equation $z^{4}-14z^{2}+1$ .
Here, the solutions for $z^{2}\;is\;7\pm{4\sqrt{3}}$ and the solution for $z\;is\;\pm\sqrt{7\pm{4\sqrt{3}}}$ .
Using what we know about the tan curve we see that $\frac{5\pi}{12}, \frac{7\pi}{12}\;are\;respectively\;\sqrt{7+4\sqrt{3}}\;and\;-\sqrt{7+4\sqrt{3}}$ and then as a matter of fact since $\tan{\frac{\pi}{12}}$ is positive then the exact value of $\tan{\frac{\pi}{12}}=\sqrt{7-4\sqrt{3}}$

=)(= · Dec 15, 2022

Ahh I see thank you

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Complex q (1 Viewer)

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