# Finding an exact expression for tan(pi/12)

• Aug 25th 2010, 04:30 AM
Glitch
Finding an exact expression for tan(pi/12)
The question:
Let $z = -2 + 2i$ and $w = -1 - \sqrt{3}i$.
i) Write z and w in polar form and thus write zw in polar form.
ii) Hence find an exact expression for $tan(\frac{\pi}{12})$

My attempt:
i) This is simple, $z = \sqrt{8}e^{\frac{3\pi}{4}i}; w = 2e^{\frac{-2\pi}{3}i} ; zw = 2\sqrt{8}e^{\frac{\pi}{12}i}$
ii) I'm not sure how to go about this. I notice that the argument of the previous answer matches that of this question. However, I do not know how to attempt it.

Any help would be great!
• Aug 25th 2010, 05:04 AM
Pim
If you visualize this, you get a triangle, one point on the origin. The vector $zw$ makes an angle of $\frac{1}{12} \pi$ with the x-axis. Therefore, if you get an $a+bi$ form of $2\sqrt{8}e^{\frac{1}{12 \pi}}$ you have $tan(\frac{1}{12 \pi}) = \frac{b}{a}$
You can get the $a+bi$ form by just multiplying $z$ and $w$ and working out the brackets.

I hope this is clear, if not, please say so.
• Aug 25th 2010, 05:07 AM
Quote:

Originally Posted by Glitch
The question:
Let $z = -2 + 2i$ and $w = -1 - \sqrt{3}i$.
i) Write z and w in polar form and thus write zw in polar form.
ii) Hence find an exact expression for $tan(\frac{\pi}{12})$

My attempt:
i) This is simple, $z = \sqrt{8}e^{\frac{3\pi}{4}i}; w = 2e^{\frac{-2\pi}{3}i} ; zw = 2\sqrt{8}e^{\frac{\pi}{12}i}$
ii) I'm not sure how to go about this. I notice that the argument of the previous answer matches that of this question. However, I do not know how to attempt it.

Any help would be great!

$\displaystyle\ tan\left(\frac{\pi}{12}\right)=\frac{sin\left(\fra c{\pi}{12}\right)}{cos\left(\frac{\pi}{12}\right)}$

$e^{(i\theta)}=cos(\theta)+isin(\theta)$

so if you multiply out

$zw=(-2+2i)(-1-\sqrt{3}i)$ you will be able to continue by comparing terms
• Aug 25th 2010, 05:16 AM
Glitch
Aha, that's interesting. Thanks!
• Aug 25th 2010, 07:08 AM
Prove It
Quote:

Originally Posted by Glitch
The question:
Let $z = -2 + 2i$ and $w = -1 - \sqrt{3}i$.
i) Write z and w in polar form and thus write zw in polar form.
ii) Hence find an exact expression for $tan(\frac{\pi}{12})$

My attempt:
i) This is simple, $z = \sqrt{8}e^{\frac{3\pi}{4}i}; w = 2e^{\frac{-2\pi}{3}i} ; zw = 2\sqrt{8}e^{\frac{\pi}{12}i}$
ii) I'm not sure how to go about this. I notice that the argument of the previous answer matches that of this question. However, I do not know how to attempt it.

Any help would be great!

You don't need to go into the Complex number system here.

Just use the angle sum identity for tangent...

$\tan{(\alpha \pm \beta)} = \frac{\tan{\alpha} \pm \tan{\beta}}{1 \mp \tan{\alpha}\tan{\beta}}$.

Here $\tan{\frac{\pi}{12}} = \tan{\left(\frac{\pi}{3} - \frac{\pi}{4}\right)}$

$= \frac{\tan{\frac{\pi}{3}} - \tan{\frac{\pi}{4}}}{1 + \tan{\frac{\pi}{3}}\tan{\frac{\pi}{4}}}$
• Aug 25th 2010, 08:19 AM