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108
4
avatar+132 
If I have $e^{17πi/60}+e^{57πi/60}$, how can I simplify that? I know $e^{i\theta}= \cos(\theta) + i \sin(\theta)$ and can add the two, and I know it leads to $cis(\frac{37π}{60})$, but I just don't know how to get there. Please help
 Mar 10, 2023
 #1
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0

To simplify this expression, we can factor out the common denominator of 60:

[e^(17pi*i)/60]  + [e^(57pi*i)/60] = (e^(17pi*i) + e^(57pi*i)) / 60

 

Now, we can use Euler's formula to write each exponential term in terms of sines and cosines:

 

e^(17pi*i) = cos(17pi) + i*sin(17pi) = -1 + 0i = -1
e^(57pi*i) = cos(57pi) + i*sin(57pi) = -1 + 0i = -1

 

Substituting these values into the expression, we get:

 

(e^(17pi*i) + e^(57pi*i)) / 60 = (-1 - 1) / 60 = -1/30

 

Therefore, the simplified expression is -1/30.

 Mar 10, 2023
 #2
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e^(17*pi*i/60) + e^(57*pi*i/60)

= 0.629 + 0.777i - 0.987 + 156i

= -0.358 + 0.933i

 Mar 10, 2023
 #3
avatar+397 
+3

Use the trig identities

\(\displaystyle \cos(A)+\cos(B)=2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \\ \text{and} \\ \displaystyle \sin(A)+\sin(B)=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right).\)

The common 

\(\displaystyle \cos\left(\frac{A-B}{2}\right)\)

term will be equal to \(\cos(\pi/3)=1/2,\)

cancelling out the 2's.

 Mar 10, 2023
 #4
avatar+132 
+1
Ahhh yes thank you so much!!!
Saphia1123  Mar 10, 2023

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