How do you use DeMoivre's Theorem to simplify #(cos0+isin0)^20#?

1 Answer
sankarankalyanam
Jun 14, 2018

#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 theta + i sin theta)]^n = r^n (cos ntheta + i sin btheta)#

#(cos 0 + i sin 0)^(20) = cos (20*0) + i sin (20*0) = 1# as

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

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