How do you use DeMoivre's Theorem to simplify ${(cos 0 + i sin 0)}^{20}$ $(cos0+isin0)^20$ ?

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How do you use DeMoivre's Theorem to simplify ${(cos 0 + i sin 0)}^{20}$ $(cos0+isin0)^20$ ?

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Answer 1 · 2018-06-14T05:18:58

${cos 0 + i sin 0}^{20} = cos (20 \cdot 0) + i sin (20 \cdot 0) = 1$ $color(indigo)(cos 0 + i sin 0)^(20) = cos (20*0) + i sin (20*0) = 1$

Explanation:

![https://www.google.com/search?q=demorvies+theorem&client=safari&hl=en-us&prmd=ivn&source=lnms&tbm=isch&sa=X&ved=0ahUKEwiYmuTisNLbAhWJwI8KHXycAGsQ_AUIESgB&biw=768&bih=922#imgrc=XMEZvta0Lgq8wM:]( useruploads.socratic.org )

By De Moivre’s theorem,

$z = {[r (cos θ + i sin θ)]}^{n} = r^{n} (cos n θ + i sin b θ)$ $z = [r(cos theta + i sin theta)]^n = r^n (cos ntheta + i sin btheta)$

${(cos 0 + i sin 0)}^{20} = cos (20 \cdot 0) + i sin (20 \cdot 0) = 1$ $(cos 0 + i sin 0)^(20) = cos (20*0) + i sin (20*0) = 1$ as

$cos 0 = 1, sin 0 = 0 & r = 1$ $cos 0 = 1, sin 0 = 0 & r = 1$

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How do you use DeMoivre's Theorem to simplify ${(cos 0 + i sin 0)}^{20}$ $(cos0+isin0)^20$ ?

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How do you use DeMoivre's Theorem to simplify ${(cos 0 + i sin 0)}^{20}$ $(cos0+isin0)^20$ ?