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

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

Guest Oct 25, 2014

#3
+91451
+10

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

$$\\=2\\ \ne 1\\ The equation is not true$$$So I agree with Alan's answer. Melody Oct 26, 2014 Sort: ### 3+0 Answers #1 +17711 +5 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 geno3141 Oct 25, 2014 #2 +91451 0 Alan's answer is here. http://web2.0calc.com/questions/cis-for-melody Alan's and Gino's answers are different. Melody Oct 26, 2014 #3 +91451 +10 Best Answer $$\\LHS\\ =(cis60)^{1991}+(cis(-60))^{1991}+1\\ =(e^{60i})^{1991}+(e^{-60i})^{1991}+1\\ =(e^{60*1991*i})+(e^{-60*{1991}*i})+1\\ =(cos(60*1991)+isin(60*1991))+(cos(-60*1991)+isin(-60*1991))+1\\ Now\;\; cos(-\theta)=cos(\theta)\;\;and\;\;sin(-\theta)=-sin(\theta)\\ so\\ =cos(60*1991)+cos(-60*1991)+isin(60*1991)+isin(-60*1991)+1\\ =cos(60*1991)+cos(-60*1991)+i[sin(60*1991)+sin(-60*1991)]+1\\ =cos(60*1991)+cos(60*1991)+i[sin(60*1991)-sin(60*1991)]+1\\ =2cos(60*1991)+i[0]+1\\ =2cos(60*1991)+1\\ =2cos(60*[6*331+5])+1\\ =2cos(60*6*331+60*5])+1\\ =2cos(360*331+60*5)+1\\ =2cos(60*5)+1\\ =2cos(300)+1\\ =2\times\frac{1}{2}+1\\ =1+1\\$$ $$\\=2\\ \ne 1\\ The equation is not true$$$

So I agree with Alan's answer.

Melody  Oct 26, 2014

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