How do you use demoivre's theorem to simplify #(cos((5pi)/4)+isin((5pi)/4))^10#?

Question

4475 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-13T12:53:43

#{cos((5pi)/4)+isin((5pi)/4)}^10=i#

According to Demoivre's theorem,

#{r(costheta+isintheta)}^n=r^n(cosntheta+isinntheta)#

Hence, #{cos((5pi)/4)+isin((5pi)/4)}^10#

= #{cos((5pi)/4xx10)+isin((5pi)/4xx10)}#

= #{cos((50pi)/4)+isin((50pi)/4)}#

= #{cos((25pi)/2)+isin((25pi)/2)}#

= #{cos(12pi+pi/2)+isin(12pi+pi/2)}#

= #{cos(pi/2)+isin(pi/2)}#

= #{0+ixx1}#

= #i#