How do you multiply $e^{\frac{π}{12} i} \cdot e^{π i}$ $e^((pi)/12i ) * e^( pi i )$ in trigonometric form?

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How do you multiply $e^{\frac{π}{12} i} \cdot e^{π i}$ $e^((pi)/12i ) * e^( pi i )$ in trigonometric form?

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Answer 1 · 2017-12-11T06:25:10

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

Explanation:

We know that

$e^{a} \cdot e^{b} = e^{a + b}$ $e^a* e^b=e^(a+b)$

$cos (a + b) = cos a cos b - sin a sin b$ $cos(a+b)=cosacosb-sinasinb$

$sin (a + b) = sin a cos b + sin b cos a$ $sin(a+b)=sinacosb+sinbcosa$

Therefore,

$e^{\frac{π}{12} i} \cdot e^{i π} = e^{\frac{13}{12} i π}$ $e^(pi/12i)*e^(ipi)=e^(13/12ipi)$

According to Euler's Identity

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

But,

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

Therefore,

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

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

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

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

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

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

So,

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

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