+0

0
26
4
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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
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
0

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
+269
+2

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
+116
+1
Ahhh yes thank you so much!!!
Saphia1123  Mar 10, 2023