How do you evaluate $e^{\frac{3 π}{2} i} - e^{\frac{π}{8} i}$ $e^( (3 pi)/2 i) - e^( ( pi)/8 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{3 π}{2} i} - e^{\frac{π}{8} i}$ $e^( (3 pi)/2 i) - e^( ( pi)/8 i)$ using trigonometric functions?

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Answer 1 · 2017-12-24T06:57:51

The answer is $= - \frac{1}{2} \sqrt{\sqrt{2} + 2} - \frac{i}{2} \sqrt{2 - \sqrt{2}} - i$ $=-1/2sqrt(sqrt2+2)-i/2sqrt(2-sqrt2)-i$

Explanation:

Apply Euler's Identity

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

$cos 2 θ = 2 {cos}^{2} θ - 1$ $cos2theta=2cos^2theta-1$ , $\Rightarrow$ $=>$ , $cos θ = \sqrt{\frac{cos 2 θ + 1}{2}}$ $costheta=sqrt((cos2theta+1)/2)$

$cos 2 θ = 1 - 2 {sin}^{2} θ$ $cos2theta=1-2sin^2theta$ , $\Rightarrow$ $=>$ , $sin θ = \sqrt{\frac{1 - cos 2 θ}{2}}$ $sintheta=sqrt((1-cos2theta)/2)$

Therefore,

$e^{\frac{3}{2} π i} - e^{\frac{1}{8} π i} = cos (\frac{3}{2} π) + i sin (\frac{3}{2} π) - cos (\frac{1}{8} π) - i sin (\frac{1}{8} π)$ $e^(3/2pii)-e^(1/8pii)=cos(3/2pi)+isin(3/2pi)-cos(1/8pi)-isin(1/8pi)$

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

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

$cos (\frac{3}{2} π) = 0$ $cos(3/2pi)=0$

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

So,

$e^{\frac{3}{2} π i} - e^{\frac{1}{8} π i} = 0 - i - \frac{1}{2} \sqrt{\sqrt{2} + 2} - i \frac{1}{2} \sqrt{2 - \sqrt{2}}$ $e^(3/2pii)-e^(1/8pii)=0-i-1/2sqrt(sqrt2+2)-i1/2sqrt(2-sqrt2)$

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How do you evaluate $e^{\frac{3 π}{2} i} - e^{\frac{π}{8} i}$ $e^( (3 pi)/2 i) - e^( ( pi)/8 i)$ using trigonometric functions?

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