How do you evaluate $e^{\frac{π}{12} i} - e^{\frac{3 π}{8} i}$ $e^( ( pi)/12 i) - e^( ( 3pi)/8 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{π}{12} i} - e^{\frac{3 π}{8} i}$ $e^( ( pi)/12 i) - e^( ( 3pi)/8 i)$ using trigonometric functions?

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Answer 1 · 2017-07-20T06:41:18

See the explanation below

Explanation:

We apply Euler's relation

$e^{i θ} = cos θ + i sin θ$ $e^(itheta)=costheta+isintheta$

$e^{\frac{1}{12} π i} = cos (\frac{1}{12} π) + i sin (\frac{1}{12} π)$ $e^(1/12pii)=cos(1/12pi)+isin(1/12pi)$

$e^{\frac{3}{8} π i} = cos (\frac{3}{8} π) + i sin (\frac{3}{8} π)$ $e^(3/8pii)=cos(3/8pi)+isin(3/8pi)$

$cos (\frac{1}{12} π) = cos (\frac{1}{3} π - \frac{1}{4} π)$ $cos(1/12pi)=cos(1/3pi-1/4pi)$

$= cos (\frac{1}{3} π) cos (\frac{1}{4} π) + sin (\frac{1}{3} π) sin (\frac{1}{4} π)$ $=cos(1/3pi)cos(1/4pi)+sin(1/3pi)sin(1/4pi)$

$= \frac{1}{2} \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}$ $=1/2*sqrt2/2+sqrt3/2*sqrt2/2$

$= \frac{\sqrt{2} + \sqrt{6}}{4}$ $=(sqrt2+sqrt6)/4$

$sin (\frac{1}{12} π) = sin (\frac{1}{3} π - \frac{1}{4} π)$ $sin(1/12pi)=sin(1/3pi-1/4pi)$

$= sin (\frac{1}{3} π) cos (\frac{1}{4} π) - sin (\frac{1}{4} π) cos (\frac{1}{3} π)$ $=sin(1/3pi)cos(1/4pi)-sin(1/4pi)cos(1/3pi)$

$= \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}$ $=sqrt3/2*sqrt2/2-sqrt2/2*1/2$

$= \frac{\sqrt{6} - \sqrt{2}}{4}$ $=(sqrt6-sqrt2)/4$

$cos (\frac{3}{4} π) = 1 - 2 {sin}^{2} (\frac{3}{8} π)$ $cos(3/4pi)=1-2sin^2(3/8pi)$

$2 {sin}^{2} (\frac{3}{8} π) = 1 - cos (\frac{3}{4} π) = 1 + \frac{\sqrt{2}}{2}$ $2sin^2(3/8pi)=1-cos(3/4pi)=1+sqrt2/2$

$sin (\frac{3}{8} π) = \sqrt{\frac{1}{2} (1 + \frac{\sqrt{2}}{2})} = \frac{1}{2} \sqrt{2 + \sqrt{2}}$ $sin(3/8pi)=sqrt(1/2(1+sqrt2/2))=1/2sqrt(2+sqrt2)$

$cos (\frac{3}{4} π) = 2 {cos}^{2} (\frac{3}{8} π) - 1$ $cos(3/4pi)=2cos^2(3/8pi)-1$

$2 {cos}^{2} (\frac{3}{8} π) = 1 + cos (\frac{3}{4} π) = 1 - \frac{\sqrt{2}}{2}$ $2cos^2(3/8pi)=1+cos(3/4pi)=1-sqrt2/2$

$cos (\frac{3}{8} π) = \frac{1}{2} \sqrt{2 - \sqrt{2}}$ $cos(3/8pi)=1/2sqrt(2-sqrt2)$

Therefore,

$e^{\frac{1}{12} π i} - e^{\frac{3}{8} π i} = cos (\frac{1}{12} π) + i sin (\frac{1}{12} π) - cos (\frac{3}{8} π) - i sin (\frac{3}{8} π)$ $e^(1/12pii)-e^(3/8pii)=cos(1/12pi)+isin(1/12pi)-cos(3/8pi)-isin(3/8pi)$

$= (\frac{\sqrt{2} + \sqrt{6}}{4}) - \frac{1}{2} \sqrt{2 - \sqrt{2}} + i ((\frac{\sqrt{6} - \sqrt{2}}{4}) - \frac{1}{2} \sqrt{2 + \sqrt{2}})$ $=((sqrt2+sqrt6)/4)-1/2sqrt(2-sqrt2)+i(((sqrt6-sqrt2)/4)-1/2sqrt(2+sqrt2))$

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How do you evaluate $e^{\frac{π}{12} i} - e^{\frac{3 π}{8} i}$ $e^( ( pi)/12 i) - e^( ( 3pi)/8 i)$ using trigonometric functions?

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