How can you use trigonometric functions to simplify $8 e^{\frac{19 π}{12} i}$ $8 e^( ( 19 pi)/12 i )$ into a non-exponential complex number?

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Answer 1 · 2017-12-04T12:54:29

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

Apply Euler's identity

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

$\frac{19}{12} π = \frac{5}{4} π + \frac{1}{3} π$ $19/12pi=5/4pi+1/3pi$

$cos (\frac{19}{12} π) = cos (\frac{5}{4} π + \frac{1}{3} π)$ $cos(19/12pi)=cos(5/4pi+1/3pi)$

$= cos (\frac{5}{4} π) cos (\frac{1}{3} π) - sin (\frac{5}{4} π) sin (\frac{1}{3} π)$ $=cos(5/4pi)cos(1/3pi)-sin(5/4pi)sin(1/3pi)$

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

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

$sin (\frac{19}{12} π) = sin (\frac{5}{4} π + \frac{1}{3} π)$ $sin(19/12pi)=sin(5/4pi+1/3pi)$

$= sin (\frac{5}{4} π) cos (\frac{1}{3} π) + sin (\frac{1}{3} π) cos (\frac{5}{4} π)$ $=sin(5/4pi)cos(1/3pi)+sin(1/3pi)cos(5/4pi)$

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

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

Therefore,

$8 e^{\frac{19}{12} π} = 8 cos (\frac{19}{12} π) + i 8 sin (\frac{19}{12} π)$ $8e^(19/12pi)=8cos(19/12pi)+i8sin(19/12pi)$

$= 8 \cdot (\frac{\sqrt{6} - \sqrt{2}}{4}) - 8 i (- \frac{\sqrt{2} + \sqrt{6}}{4})$ $=8*((sqrt6-sqrt2)/4)-8i(-(sqrt2+sqrt6)/4)$

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

How can you use trigonometric functions to simplify 8 e^( ( 19 pi)/12 i ) 8e19π12i 8 e^( ( 19 pi)/12 i ) into a non-exponential complex number?