Question

Use De Moivre's theorem to simplify the expression. Put in form a+bi, and round to the nearest tenth.
[cos(pi/3)+i*sin(pi/3)]^6

Answer 1

multiply the angle by 6, then evaluate the trig funcions

Answer 2

\[(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta)\]\[(\cos(\frac\pi3)+i\sin(\frac\pi3))^6=\cos(6\cdot\frac\pi3)+i\sin(6\cdot\frac\pi3)\]

Answer 3

\[(\cos(\frac{\pi}{3})+i\sin)\frac{\pi}{3}))^6\]
\[\cos(2\pi)+i\sin(2\pi)\]
\[1+0i\]
\[1\]

Answer 4

Thank you, but what I don't understand is how you go from \[6\pi/3 \to 2\pi?\]

Answer 5

nevermind, simple mistake lol:) thank you