How can you use trigonometric functions to simplify $5 e^{\frac{13 π}{12} i}$ $5 e^( ( 13 pi)/12 i )$ into a non-exponential complex number?

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How can you use trigonometric functions to simplify $5 e^{\frac{13 π}{12} i}$ $5 e^( ( 13 pi)/12 i )$ into a non-exponential complex number?

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Answer 1 · 2017-08-11T09:09:51

The answer is $= - \frac{5}{4} (\sqrt{6} + \sqrt{2}) + i \frac{5}{4} (\sqrt{2} - \sqrt{6})$ $=-5/4(sqrt6+sqrt2)+i5/4(sqrt2-sqrt6)$

Explanation:

We apply Euler's relation

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

$5 e^{\frac{13}{12} i π} = 5 (cos (\frac{13}{12} π) + i sin (\frac{13}{12} π))$ $5e^(13/12ipi)=5(cos(13/12pi)+isin(13/12pi))$

$cos (\frac{13}{12} π) = cos (\frac{1}{3} π + \frac{3}{4} π)$ $cos(13/12pi)=cos(1/3pi+3/4pi)$

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

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

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

$sin (\frac{13}{12} π) = sin (\frac{1}{3} π + \frac{3}{4} π)$ $sin(13/12pi)=sin(1/3pi+3/4pi)$

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

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

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

Therefore,

$5 e^{\frac{13}{12} i π} = - \frac{5}{4} (\sqrt{6} + \sqrt{2}) + i \frac{5}{4} (\sqrt{2} - \sqrt{6})$ $5e^(13/12ipi)=-5/4(sqrt6+sqrt2)+i5/4(sqrt2-sqrt6)$

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How can you use trigonometric functions to simplify $5 e^{\frac{13 π}{12} i}$ $5 e^( ( 13 pi)/12 i )$ into a non-exponential complex number?

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How can you use trigonometric functions to simplify 5e13π12i 5 e^( ( 13 pi)/12 i ) into a non-exponential complex number?

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How can you use trigonometric functions to simplify $5 e^{\frac{13 π}{12} i}$ $5 e^( ( 13 pi)/12 i )$ into a non-exponential complex number?