How do you evaluate $e^{\frac{7 π}{6} i} - e^{\frac{14 π}{3} i}$ $e^( ( 7 pi)/6 i) - e^( ( 14 pi)/3 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{7 π}{6} i} - e^{\frac{14 π}{3} i}$ $e^( ( 7 pi)/6 i) - e^( ( 14 pi)/3 i)$ using trigonometric functions?

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Answer 1 · 2016-03-30T15:33:53

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

Explanation:

$e^{\frac{7 π i}{6}} = cos (\frac{7 π}{6}) + i sin (\frac{7 π}{6})$ $e^((7pi i)/6)= cos ((7pi)/6) + i sin ((7pi)/6)$

= $- \frac{cos π}{6} - i \frac{sin π}{6}$ $-cos pi/6 - i sin pi/6$ $[\frac{7 π}{6} = π + \frac{π}{6}$ $[(7pi)/6 = pi +pi/6$ , In the third quadrant both sin and cos would b negative]

Like wise $e^{\frac{14 π i}{3}} = cos (\frac{14 π}{3}) + i sin (\frac{14 π}{3})$ $e^((14pi i)/3)= cos ((14pi)/3) + i sin ((14pi)/3)$

= $cos (\frac{2 π}{3}) + i sin (\frac{2 π}{3})$ $cos ((2pi)/3) + i sin ((2pi)/3)$ $[\frac{14 π}{3} = 4 π + \frac{2 π}{3} = \frac{2 π}{3}]$ $[(14pi)/3= 4pi +(2pi)/3= (2pi)/3]$

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

Now combining both expressions, it would be $- \frac{\sqrt{3}}{2} - \frac{i}{2} + \frac{1}{2} - i \frac{\sqrt{3}}{2}$ $-sqrt3 /2 -i/2 +1/2 -i sqrt3 /2$

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How do you evaluate $e^{\frac{7 π}{6} i} - e^{\frac{14 π}{3} i}$ $e^( ( 7 pi)/6 i) - e^( ( 14 pi)/3 i)$ using trigonometric functions?

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Explanation:

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How do you evaluate e^( ( 7 pi)/6 i) - e^( ( 14 pi)/3 i)e7π6i−e14π3i e^( ( 7 pi)/6 i) - e^( ( 14 pi)/3 i) using trigonometric functions?

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How do you evaluate $e^{\frac{7 π}{6} i} - e^{\frac{14 π}{3} i}$ $e^( ( 7 pi)/6 i) - e^( ( 14 pi)/3 i)$ using trigonometric functions?