(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.