How can you use trigonometric functions to simplify $15 e^{\frac{17 π}{12} i}$ $15 e^( ( 17 pi)/12 i )$ into a non-exponential complex number?

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How can you use trigonometric functions to simplify $15 e^{\frac{17 π}{12} i}$ $15 e^( ( 17 pi)/12 i )$ into a non-exponential complex number?

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Answer 1 · 2018-08-04T09:15:44

The answer is $= \frac{15}{4} ((\sqrt{2} - \sqrt{6}) - i (\sqrt{2} + \sqrt{6}))$ $=15/4((sqrt2-sqrt6)-i(sqrt2+sqrt6))$

Explanation:

Apply the identity

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

And

$\frac{17}{12} π = \frac{3}{4} π + \frac{2}{3} π$ $17/12pi=3/4pi+2/3pi$

Therefore,

$15 e^{\frac{17}{12} i π} = 15 (cos (\frac{3}{4} π + \frac{2}{3} π) + i sin (\frac{3}{4} π + \frac{2}{3} π))$ $15e^(17/12ipi)=15(cos(3/4pi+2/3pi)+isin(3/4pi+2/3pi))$

$cos (\frac{3}{4} π + \frac{2}{3} π) = cos (\frac{3}{4} π) cos (\frac{2}{3} π) - sin (\frac{3}{4} π) sin (\frac{2}{3} π)$ $cos(3/4pi+2/3pi)=cos(3/4pi)cos(2/3pi)-sin(3/4pi)sin(2/3pi)$

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

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

$sin (\frac{3}{4} π + \frac{2}{3} π) = sin (\frac{3}{4} π) cos (\frac{2}{3} π) + cos (\frac{3}{4} π) sin (\frac{2}{3} π)$ $sin(3/4pi+2/3pi)=sin(3/4pi)cos(2/3pi)+cos(3/4pi)sin(2/3pi)$

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

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

Finally,

$15 e^{\frac{17}{12} i π} = \frac{15}{4} ((\sqrt{2} - \sqrt{6}) + i (- \sqrt{2} - \sqrt{6}))$ $15e^(17/12ipi)=15/4((sqrt2-sqrt6)+i(-sqrt2-sqrt6))$

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How can you use trigonometric functions to simplify $15 e^{\frac{17 π}{12} i}$ $15 e^( ( 17 pi)/12 i )$ into a non-exponential complex number?

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