We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
723
3
avatar

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

 Oct 25, 2014

Best Answer 

 #3
avatar+102430 
+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.    

 Oct 26, 2014
 #1
avatar+17776 
+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

 Oct 25, 2014
 #2
avatar+102430 
0

Alan's answer is here.

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

Alan's and Gino's answers are different.   

 Oct 26, 2014
 #3
avatar+102430 
+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

17 Online Users

avatar
avatar