(1cis60)^1991+(1cis(-60))^1991+1=1 i need to prove this but the calculator wont give me the exact answers

$$\\LHS\\
=(cis60)^{1991}+(cis(-60))^{1991}+1\\
=(e^{60i})^{1991}+(e^{-60i})^{1991}+1\\
=(e^{60*1991*i})+(e^{-60*{1991}*i})+1\\
=(cos(60*1991)+isin(60*1991))+(cos(-60*1991)+isin(-60*1991))+1\\
Now\;\; cos(-\theta)=cos(\theta)\;\;and\;\;sin(-\theta)=-sin(\theta)\\
so\\
=cos(60*1991)+cos(-60*1991)+isin(60*1991)+isin(-60*1991)+1\\
=cos(60*1991)+cos(-60*1991)+i[sin(60*1991)+sin(-60*1991)]+1\\
=cos(60*1991)+cos(60*1991)+i[sin(60*1991)-sin(60*1991)]+1\\
=2cos(60*1991)+i[0]+1\\
=2cos(60*1991)+1\\
=2cos(60*[6*331+5])+1\\
=2cos(60*6*331+60*5])+1\\
=2cos(360*331+60*5)+1\\
=2cos(60*5)+1\\
=2cos(300)+1\\
=2\times\frac{1}{2}+1\\
=1+1\\$$

$$\\=2\\
\ne 1\\
$The equation is not true$$$

So I agree with Alan's answer.    

Using deMoivre's formula:

(1cis60)^1991  =  (1^1991)cis(60*1991)  =  cis(119460)  =  cis(300)  =  .5 + -√(3)/2i

(1cis(-60))^1991  =  (1^1991)cis(-60*1991)  =  cis(-119460)  =  cis(-300)  =  .5 + √(3)/2i

Adding:   .5 + -√(3)/2i  +  .5 + √(3)/2i  = 1 + 0i  =  1

Alan's answer is here.

http://web2.0calc.com/questions/cis-for-melody

Alan's and Gino's answers are different.