So - the following http://prntscr.com/b2mq1o

1) Use DeMoivre's Theorem.

so... it should be

sqrt3^4 (cos4 * 5pi/6 + i sin4 * 5pi/6)

after that I simplify, but that's where I get confused.

9(cos20pi/6 + i sin 20pi/6)

Question

So - the following http://prntscr.com/b2mq1o

1) Use DeMoivre's Theorem. 

so... it should be 

sqrt3^4 (cos4 * 5pi/6 + i sin4 * 5pi/6)

after that I simplify, but that's where I get confused. 


9(cos20pi/6 + i sin 20pi/6)

baru · Answer

we know that $\theta + 2\pi = \theta$

so using mixed fractions, re-write 20pi/6 as

$$3\pi + \frac{1}{3} \pi \=2\pi + \pi + \frac{1}{3}\pi \=\pi + \frac{\pi}{3}$$

baru · Answer

this corresponds to an angle of pi/3 in the 'third quadrant'|dw:1462938632945:dw|

baru · Answer

can you take it from here?

sloppycanada · Answer

Yes, from there it's 

9(-cos4pi/3) + i(-sin4pi/3)

baru · Answer

cos pi/3 = ?
sin pi/3 = ?

baru · Answer

ohh.. sorry i've made a mistake

baru · Answer

it simplifies to
$$9(-\cos(\pi/3) - i \sin(\pi/3))$$

sloppycanada · Answer

Then what?

baru · Answer

substitute the values for cos(pi/3) and sin(pi/3)

sloppycanada · Answer

What values?

baru · Answer

do you know the value of sin(60) ?

sloppycanada · Answer

The answer is C, right?

baru · Answer

yea

baru · Answer

http://study.com/cimages/multimages/16/imagetrigfunctions9.jpg

baru · Answer

you need to memorize that table

sloppycanada · Answer

Really..? I'm horrible at memorizing things.

baru · Answer

sorry, no way around it :P