How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{π}{6} i}$ $e^( ( pi)/4 i) - e^( ( pi)/6 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{π}{6} i}$ $e^( ( pi)/4 i) - e^( ( pi)/6 i)$ using trigonometric functions?

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Answer 1 · 2016-05-18T09:40:26

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

Explanation:

$e^{\frac{π}{4} i} = cos (\frac{π}{4}) + i sin (\frac{π}{4})$ $e^(pi/4i)=cos(pi/4)+isin(pi/4)$ and $e^{\frac{π}{6} i} = cos (\frac{π}{6}) + i sin (\frac{π}{6})$ $e^(pi/6i)=cos(pi/6)+isin(pi/6)$

Hence, $e^{\frac{π}{4} i} - e^{\frac{π}{6} i} = cos (\frac{π}{4}) + i sin (\frac{π}{4}) - cos (\frac{π}{6}) - i sin (\frac{π}{6})$ $e^(pi/4i)-e^(pi/6i)=cos(pi/4)+isin(pi/4)-cos(pi/6)-isin(pi/6)$

= $cos (\frac{π}{4}) - cos (\frac{π}{6}) + i (sin (\frac{π}{4}) - sin (\frac{π}{6}))$ $cos(pi/4)-cos(pi/6)+i(sin(pi/4)-sin(pi/6))$

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

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

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How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{π}{6} i}$ $e^( ( pi)/4 i) - e^( ( pi)/6 i)$ using trigonometric functions?

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