How do you use DeMoivre's Theorem to simplify ${(5 (cos 20 + i sin 20))}^{3}$ $(5(cos20+isin20))^3$ ?

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Answer 1 · 2016-11-19T17:42:14

The answer is $= \frac{125}{2} (1 + i \sqrt{3})$ $=125/2(1+isqrt3)$

DeMoivre 'theorem is

${(cos θ + i sin θ)}^{n} = cos n θ + i sin n θ$ $(costheta+isintheta)^n= cosntheta+isinntheta$

Therefore,

${(5 (cos 20 + i sin 20))}^{3}$ $(5(cos20+isin20))^3$

$= 5^{3} (cos 3 \cdot 20 + i sin 3 \cdot 20)$ $=5^3(cos3*20+isin3*20)$

$= 125 (cos 60 + i sin 60)$ $=125 (cos60+isin60)$

$= 125 (\frac{1}{2} + i \frac{\sqrt{3}}{2})$ $=125(1/2+isqrt3/2)$

How do you use DeMoivre's Theorem to simplify (5(cos20+isin20))3(5(cos20+isin20))^3?