This is the answer I ended up with:
=(-2)[cos(-pi/6)+isin(-pi/6)]
Is this right?
Does it evaluate to the same decimal (or exact) form?
. . . . .[32(cos(2π)+isin(2π))][3(cos(32π)+isin(32π))]
. . . . . . . . . .\(\displaystyle =\, \left[\, \dfrac{2}{3}\, (0\, +\, 1i)\, \right]\, \left[\, 3\, \left(-\dfrac{1}{2}\, +\, \dfrac{\sqrt{\strut 3\,}}{2}\, i\right)\, \right]\)
. . . . . . . . . .\(\displaystyle =\, \left[\, \dfrac{2}{3}\, i\, \right]\, \left[\, -\dfrac{3}{2}\, +\, \dfrac{3\, \sqrt{\strut 3\,}}{2}\, i\, \right]\)
. . . . . . . . . .\(\displaystyle =\,-i\, -\, \sqrt{\strut 3\, }\, =\, -\sqrt{\strut 3\,}\, -\, i\)
Your proposed solution, on the other hand:
. . . . .(−2)[cos(−6π)+isin(−6π)]
. . . . . . . . . .\(\displaystyle =\, (-2)\, \left[\dfrac{\sqrt{\strut 3\,}}{2}\, -\, \dfrac{1}{2}\, i\right]\)
. . . . . . . . . .\(\displaystyle =\, -\sqrt{\strut 3\,}\, +\, i\)
How did you obtain your solution? Please show
all of your steps. Thank you!
