4^6=4096
115*6=690
cos(690)=sqrt(3)/2
sin(690)=-1/2
Find
[4(cos115+isin115)]^6
3 answers
De Moivre's Theorem:
(r(cosØ + i sinØ))^n
= r^n(cos(nØ) + i sin(nØ))
in our case ....
[4(cos115+isin115)]^6
= 4^6( cos 690 + i sin 690)
I will assume my meant degrees, since the numbers work out nicely in degrees.
Now 690° is coterminal with 330° or -30°
sin(-30°) = -1/2, and cos(-30) = √3/2
so continuing...
= 4096(√3/2 - (1/2)i)
= 2048√3 - 2048
(r(cosØ + i sinØ))^n
= r^n(cos(nØ) + i sin(nØ))
in our case ....
[4(cos115+isin115)]^6
= 4^6( cos 690 + i sin 690)
I will assume my meant degrees, since the numbers work out nicely in degrees.
Now 690° is coterminal with 330° or -30°
sin(-30°) = -1/2, and cos(-30) = √3/2
so continuing...
= 4096(√3/2 - (1/2)i)
= 2048√3 - 2048
oops, left out the i in the last line
= 2048√3 - 2048i
(I could always claim that you should have "imagined" it)
= 2048√3 - 2048i
(I could always claim that you should have "imagined" it)