How do you evaluate $e^{\frac{13 π}{8} i} - e^{\frac{5 π}{4} i}$ $e^( ( 13 pi)/8 i) - e^( ( 5 pi)/4 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{13 π}{8} i} - e^{\frac{5 π}{4} i}$ $e^( ( 13 pi)/8 i) - e^( ( 5 pi)/4 i)$ using trigonometric functions?

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Answer 1 · 2017-12-18T11:09:33

Euler's theorem: $e^{i θ} = cos θ - i sin θ$ $e^(itheta)=costheta-isintheta$

Explanation:

$e^{\frac{13 π}{8} i} - e^{\frac{5 π}{4} i}$ $e^((13pi)/8i)-e^((5pi)/4i)$
$= (cos (\frac{13 π}{8}) + i sin (\frac{13 π}{8})) - (cos (\frac{5 π}{4}) + i sin (\frac{5 π}{4}))$ $=(cos((13pi)/8)+isin((13pi)/8))-(cos((5pi)/4)+isin((5pi)/4))$
$= sin (\frac{π}{8}) - i cos (\frac{π}{8}) - \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i$ $=sin(pi/8)-icos(pi/8)-1/sqrt(2)+1/sqrt(2)i$
$≅ - 0.32442 - 0.21677 i$ $~=-0.32442 - 0.21677 i$

Answer 2 · 2017-12-18T13:15:45

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

Explanation:

Reminder :

Euler's relation

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

Here, we have

$z = e^{\frac{13}{8} π} - e^{\frac{5}{4} π} = cos (\frac{13}{8} π) + i sin (\frac{13}{8} π) - cos (\frac{5}{4} π) - i sin (\frac{5}{4} π)$ $z=e^(13/8pi)-e^(5/4pi)=cos(13/8pi)+isin(13/8pi)-cos(5/4pi)-isin(5/4pi)$

Therefore,

$\frac{13}{8} π = \frac{5}{8} π + π = \frac{1}{8} π + \frac{3}{2} π$ $13/8pi=5/8pi+pi=1/8pi+3/2pi$

$\frac{5}{4} π = \frac{1}{4} π + π$ $5/4pi=1/4pi+pi$

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

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

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

So,
$z = cos (\frac{1}{8} π + \frac{3}{2} π) + i sin (\frac{1}{8} π + \frac{3}{2} π) - cos (\frac{1}{4} π + π) - i sin (\frac{1}{4} π + π)$ $z=cos(1/8pi+3/2pi)+isin(1/8pi+3/2pi)-cos(1/4pi+pi)-isin(1/4pi+pi)$

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

$= \frac{\sqrt{2 + \sqrt{2}}}{2} \cdot 0 - \frac{\sqrt{2 - \sqrt{2}}}{2} \cdot (- 1) = \frac{\sqrt{2 - \sqrt{2}}}{2}$ $=(sqrt(2+sqrt2))/2*0-(sqrt(2-sqrt2))/2*(-1)=(sqrt(2-sqrt2))/2$

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

$= \frac{\sqrt{2 - \sqrt{2}}}{2} \cdot 0 + \frac{\sqrt{2 + \sqrt{2}}}{2} \cdot (- 1) = - \frac{\sqrt{2 + \sqrt{2}}}{2}$ $=(sqrt(2-sqrt2))/2*0+(sqrt(2+sqrt2))/2*(-1)=-(sqrt(2+sqrt2))/2$

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

$= \frac{\sqrt{2}}{2} \cdot - 1 - \frac{\sqrt{2}}{2} \cdot 0 = - \frac{\sqrt{2}}{2}$ $=sqrt2/2*-1-sqrt2/2*0=-sqrt2/2$

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

$= \frac{\sqrt{2}}{2} \cdot - 1 + \frac{\sqrt{2}}{2} \cdot 0 = \frac{\sqrt{2}}{2}$ $=sqrt2/2*-1+sqrt2/2*0=sqrt2/2$

Finally,

$z = \frac{\sqrt{2 - \sqrt{2}}}{2} - i \frac{\sqrt{2 + \sqrt{2}}}{2} + \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}$ $z=(sqrt(2-sqrt2))/2-i(sqrt(2+sqrt2))/2+sqrt2/2-isqrt2/2$

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

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How do you evaluate $e^{\frac{13 π}{8} i} - e^{\frac{5 π}{4} i}$ $e^( ( 13 pi)/8 i) - e^( ( 5 pi)/4 i)$ using trigonometric functions?

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