How do you evaluate $e^{\frac{11 π}{6} i} - e^{\frac{9 π}{8} i}$ $e^( ( 11 pi)/6 i) - e^( ( 9 pi)/8 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{11 π}{6} i} - e^{\frac{9 π}{8} i}$ $e^( ( 11 pi)/6 i) - e^( ( 9 pi)/8 i)$ using trigonometric functions?

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Answer 1 · 2016-04-09T05:37:47

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

Explanation:

$e^{i x} = cos x + i sin x$ $e^(ix)=cos x + i sinx$ .
$cos (2 π - x) = cos x, sin (2 π - x) = - sin x, cos (π + x) = - cos x and sin (π + x) = - sin x$ $cos(2pi-x)=cos x, sin (2pi-x)=-sin x, cos(pi+x)=-cos x and sin(pi+x)=-sin x$ .

So, $e^{11 \frac{π}{6} i} = cos (11 \frac{π}{6}) + i sin (11 \frac{π}{6}) = cos (2 π - \frac{π}{6}) + i sin (2 π - \frac{π}{6}) = cos (\frac{π}{6}) - i sin (\frac{π}{6}) = \frac{\sqrt{3}}{2} - i (\frac{1}{2})$ $e^(11pi/6i)=cos (11pi/6)+i sin(11pi/6)=cos(2pi-pi/6)+i sin(2pi-pi/6)=cos (pi/6)-i sin (pi/6)=sqrt3/2-i(1/2)$ .
$e^{9 \frac{π}{8}} = cos (9 \frac{π}{8}) + i sin (9 \frac{π}{8}) = cos (π + \frac{π}{8}) + i sin (π + \frac{π}{8}) = - cos (\frac{π}{8}) - i sin (\frac{π}{8})$ $e^(9pi/8)=cos (9pi/8)+i sin(9pi/8)=cos(pi+pi/8)+i sin(pi+pi/8)=-cos (pi/8)-i sin( pi/8)$ .
$cos A = \sqrt{\frac{1 + cos 2 A}{2}}, sin A = \sqrt{\frac{1 - cos 2 A}{2}}$ $cos A = sqrt((1+cos 2A)/2), sin A=sqrt((1-cos 2A)/2)$

So, $cos (\frac{π}{8}) = \sqrt{\frac{1 + cos (\frac{π}{4})}{2}} = \sqrt{\frac{\sqrt{2} + 1}{2 \sqrt{2}}}$ $cos(pi/8)=sqrt((1+cos (pi/4))/2) = sqrt((sqrt2+1)/(2sqrt2))$ .
$sin (\frac{π}{8}) = \sqrt{\frac{1 - cos (\frac{π}{4})}{2}} = \sqrt{\frac{\sqrt{2} - 1}{2 \sqrt{2}}}$ $sin(pi/8)=sqrt((1-cos (pi/4))/2) = sqrt((sqrt2-1)/(2sqrt2))$ .

Npw, $e^{9 \frac{π}{8}} = \sqrt{\frac{\sqrt{2} + 1}{2 \sqrt{2}}} - i \sqrt{\frac{\sqrt{2} - 1}{2 \sqrt{2}}}$ $e^(9pi/8)=sqrt((sqrt2+1)/(2sqrt2))-i sqrt((sqrt2-1)/(2sqrt2))$ .

So, the given expression = $(\frac{\sqrt{3}}{2} + \sqrt{\frac{\sqrt{2} + 1}{2 \sqrt{2}}}) - i (\frac{1}{2} - \sqrt{\frac{\sqrt{2} - 1}{2 \sqrt{2}}})$ $(sqrt3/2+sqrt((sqrt2+1)/(2sqrt2)))-i(1/2-sqrt((sqrt2-1)/(2sqrt2)))$

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How do you evaluate $e^{\frac{11 π}{6} i} - e^{\frac{9 π}{8} i}$ $e^( ( 11 pi)/6 i) - e^( ( 9 pi)/8 i)$ using trigonometric functions?

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How do you evaluate e^( ( 11 pi)/6 i) - e^( ( 9 pi)/8 i)e11π6i−e9π8i e^( ( 11 pi)/6 i) - e^( ( 9 pi)/8 i) using trigonometric functions?

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How do you evaluate $e^{\frac{11 π}{6} i} - e^{\frac{9 π}{8} i}$ $e^( ( 11 pi)/6 i) - e^( ( 9 pi)/8 i)$ using trigonometric functions?