How do you evaluate $e^{\frac{7 π}{4} i} - e^{\frac{5 π}{3} i}$ $e^( ( 7 pi)/4 i) - e^( ( 5 pi)/3 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{7 π}{4} i} - e^{\frac{5 π}{3} i}$ $e^( ( 7 pi)/4 i) - e^( ( 5 pi)/3 i)$ using trigonometric functions?

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Answer 1 · 2017-08-19T07:03:45

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

Explanation:

We apply Euler's formula

$e^{i x} = cos x + i sin x$ $e^(i x)=cosx +i sinx$

So,

$e^{\frac{7}{4} π i} - e^{\frac{5}{3} π i} =$ $e^(7/4pi i)-e^(5/3pi i)=$

$cos (\frac{7}{4} π) + i sin (\frac{7}{4} π) - cos (\frac{5}{3} π) - i sin (\frac{5}{3} π)$ $cos(7/4pi)+isin(7/4pi)-cos(5/3pi)-isin(5/3pi)$

We calculate separately

$cos (\frac{7}{4} π) = cos (π + \frac{3}{4} π) = cos (π) cos (\frac{3}{4} π) - sin (π) sin (\frac{3}{4} π)$ $cos(7/4pi)=cos(pi+3/4pi)=cos(pi)cos(3/4pi)-sin(pi)sin(3/4pi)$

$= - 1 \cdot - \frac{\sqrt{2}}{2} - 0 = \frac{\sqrt{2}}{2}$ $=-1*-sqrt2/2-0=sqrt2/2$

$sin (\frac{7}{4} π) = sin (π + \frac{3}{4} π) = sin (π) cos (\frac{3}{4} π) + sin (\frac{3}{4} π) cos (π)$ $sin(7/4pi)=sin(pi+3/4pi)=sin(pi)cos(3/4pi)+sin(3/4pi)cos(pi)$

$= 0 + \frac{\sqrt{2}}{2} \cdot - 1 = - \frac{\sqrt{2}}{2}$ $=0+sqrt2/2*-1=-sqrt2/2$

$cos (\frac{5}{3} π) = cos (π + \frac{2}{3} π) = cos (π) cos (\frac{2}{3} π) - sin (π) sin (\frac{2}{3} π)$ $cos(5/3pi)=cos(pi+2/3pi)=cos(pi)cos(2/3pi)-sin(pi)sin(2/3pi)$

$= - 1 \cdot - \frac{\sqrt{3}}{2} - 0 = \frac{\sqrt{3}}{2}$ $=-1*-sqrt3/2-0=sqrt3/2$

$sin (\frac{5}{3} π) = sin (π + \frac{2}{3} π) = sin (π) cos (\frac{2}{3} π) + sin (\frac{2}{3} π) cos (π)$ $sin(5/3pi)=sin(pi+2/3pi)=sin(pi)cos(2/3pi)+sin(2/3pi)cos(pi)$

$= 0 + \frac{\sqrt{3}}{2} \cdot - 1 = - \frac{\sqrt{3}}{2}$ $=0+sqrt3/2*-1=-sqrt3/2$

$e^{\frac{7}{4} π i} - e^{\frac{5}{3} π i} = (\frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2}) - i (\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2})$ $e^(7/4pi i)-e^(5/3pi i)=(sqrt2/2-sqrt3/2)-i(sqrt2/2+sqrt3/2)$

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