How do you evaluate $e^{\frac{π}{2} i} - e^{\frac{23 π}{12} i}$ $e^( ( pi)/2 i) - e^( ( 23 pi)/12 i)$ using trigonometric functions?

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Answer 1 · 2017-09-13T12:04:13

$e^{i \frac{π}{2}} - e^{i \frac{23 \cdot π}{12}} = - 0.9659 + 1.2588 i$ $e^(i pi/2) - e^(i (23*pi)/12) = -0.9659 + 1.2588i$

We know $e^{i θ} = cos θ + i sin θ$ $e^(itheta) = cos theta +i sin theta$

$e^{i \frac{π}{2}} = cos (\frac{π}{2}) + i sin (\frac{π}{2}) = 0 + i = i$ $e^(i pi/2) = cos (pi/2) +i sin (pi/2) = 0 + i= i$ and

$e^{i \frac{23 \cdot π}{12}} = cos (\frac{23 π}{12}) + i sin (\frac{23 π}{12})$ $e^(i (23*pi)/12) = cos( (23pi)/12) +i sin ( (23pi)/12)$ or

$e^{i \frac{23 \cdot π}{12}} = cos 345 + i sin 345 = 0.9659 - 0.2588 i$ $e^(i (23*pi)/12) = cos 345 +i sin 345 = 0.9659 - 0.2588i$

$e^{i \frac{π}{2}} - e^{i \frac{23 \cdot π}{12}} = i - (0.9659 - 0.2588 i)$ $e^(i pi/2) - e^(i (23*pi)/12) = i- (0.9659 - 0.2588i)$ or

$e^{i \frac{π}{2}} - e^{i \frac{23 \cdot π}{12}} = - 0.9659 + 1.2588 i$ $e^(i pi/2) - e^(i (23*pi)/12) = -0.9659 + 1.2588i$ [Ans]

How do you evaluate eπ2i−e23π12i e^( ( pi)/2 i) - e^( ( 23 pi)/12 i) using trigonometric functions?