How do you evaluate $e^{\frac{π}{12} i} - e^{\frac{9 π}{8} i}$ $e^( ( pi)/12 i) - e^( ( 9 pi)/8 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{π}{12} i} - e^{\frac{9 π}{8} i}$ $e^( ( pi)/12 i) - e^( ( 9 pi)/8 i)$ using trigonometric functions?

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Answer 1 · 2017-07-16T07:29:09

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

Explanation:

We apply Euler's relation

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

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

$e^{\frac{9}{8} π i} = cos (\frac{9}{8} π) + i sin (\frac{9}{8} π)$ $e^(9/8pii)=cos(9/8pi)+isin(9/8pi)$

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

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

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

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

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

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

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

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

$cos (\frac{9}{8} π) = cos (π + \frac{1}{8} π)$ $cos(9/8pi)=cos(pi+1/8pi)$

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

$= - 1 \cdot cos (\frac{1}{8} π) - 0 \cdot cos (\frac{1}{8} π)$ $=-1*cos(1/8pi)-0*cos(1/8pi)$

$= - cos (\frac{1}{8} π)$ $=-cos(1/8pi)$

$cos (2 x) = 2 {cos}^{2} x - 1$ $Cos(2x)=2cos^2x-1$

${cos}^{2} x = \frac{1 + cos (2 x)}{2}$ $cos^2x=(1+cos(2x))/2$

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

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

$cos (\frac{9}{8} π) = - \frac{\sqrt{2 + \sqrt{2}}}{2}$ $cos(9/8pi)=-sqrt(2+sqrt2)/2$

$sin (\frac{9}{8} π) = sin (π + \frac{1}{8} π) = sin (π) cos (\frac{1}{8} π) + sin (\frac{1}{8}) π cos (π)$ $sin(9/8pi)=sin(pi+1/8pi)=sin(pi)cos(1/8pi)+sin(1/8)picos(pi)$

$= 0 - sin (\frac{1}{8} π)$ $=0-sin(1/8pi)$

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

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

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

$sin (\frac{9}{8} π) = - \frac{\sqrt{2 - \sqrt{2}}}{2}$ $sin(9/8pi)=-sqrt(2-sqrt2)/2$

Therefore,

$e^{\frac{1}{12} π i} - e^{\frac{9}{8} π i} = cos (\frac{1}{12} π) + i sin (\frac{1}{12} π) - cos (\frac{9}{8} π) + i sin (\frac{9}{8} π)$ $e^(1/12pii)-e^(9/8pii)=cos(1/12pi)+isin(1/12pi)-cos(9/8pi)+isin(9/8pi)$

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

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