Question

use de moivre's theorem to write the complex number in trigonometric form (cos(2pi/7)+isin(2pi/7))^5

Answer 1

By DeMoivre's theorem, you have

$\left(\cos\dfrac{2\pi}7+i\sin\dfrac{2\pi}7\right)^5=\cos\dfrac{10\pi}7+i\sin\dfrac{10\pi}7$

Answer 2

Answer:

$(\cos{\frac{2\pi}{7}}+i\sin{\frac{2\pi}{7}})^5 =\cos{(\frac{10\pi}{7})}+i\sin{(\frac{10\pi}{7})}$

Step-by-step explanation:

Given the complex number

$(\cos{\frac{2\pi}{7}}+i\sin{\frac{2\pi}{7}})^5$

we have to write the complex number in trigonometric form using de moivre's theorem.

By de Moivre's formula,

$(\cos x+i \sin x)^ n=\cos{nx}+i\sin{nx}$

∴ $(\cos{\frac{2\pi}{7}}+i\sin{\frac{2\pi}{7}})^5=\cos{5(\frac{2\pi}{7})}+i\sin{5(\frac{2\pi}{7})}$

$=\cos{(\frac{10\pi}{7})}+i\sin{(\frac{10\pi}{7})}$

which is required form