Polar form

• Apr 6th 2011, 04:09 AM
Rine198
Polar form
Let z=2exp(-Pi/6) and w=sqrt(2)exp(Pi/4)

Find zw in polar form and hence express tan(Pi/12) in surd form

So I manage to find zw in polar form, which is 2sqrt(2)exp(Pi/12), but I don't know how to express tan(Pi/12) in surd form. Please help me thanks.
• Apr 6th 2011, 04:38 AM
Plato
Quote:

Originally Posted by Rine198
Let z=2exp(-Pi/6) and w=sqrt(2)exp(Pi/4)
Find zw in polar form and hence express tan(Pi/12) in surd form

Do you know a formula for $\displaystyle \tan(\frac{\theta}{2})}~?$
Note that $\displaystyle \frac{\pi}{12}=\frac{1}{2}\frac{\pi}{6}$.
• Apr 6th 2011, 04:40 AM
Rine198
do u mean the t-formula?
• Apr 6th 2011, 04:58 AM
Soroban
Hello, Rine198!

Quote:

$\displaystyle \text{Express }\tan\frac{\pi}{12}\text{ in surd form.}$

I'm not sure about the "hence".
We can get the answer with some old=fashioned trig.

$\displaystyle \tan\frac{\pi}{12} \;=\;\tan\left(\frac{\pi}{4}-\frac{\pi}{6}\right) \;=\;\dfrac{\tan\frac{\pi}{4} - \tan\frac{\pi}{6}}{1 + \tan\frac{\pi}{4}\tan\frac{\pi}{5}} \;=\;\dfrac{1-\frac{1}{\sqrt{3}}} {1 + \frac{1}{\sqrt{3}}}$

$\displaystyle \text{Multiply by }\frac{\sqrt{3}}{\sqrt{3}}\!: \;\;\dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}$

$\displaystyle \text{Rationalize: }\;\dfrac{\sqrt{3} - 1}{\sqrt{3} - 1}\cdot \dfrac{\sqrt{3} - 1}{\sqrt{3}+1} \;=\;\dfrac{3 - 2\sqrt{3} + 1}{3-1}$

. . . . . . . . $\displaystyle =\;\dfrac{4 - 2\sqrt{3}}{2} \;=\;2 - \sqrt{3}$