How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{7 π}{6} i}$ $e^( ( pi)/4 i) - e^( ( 7 pi)/6 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{7 π}{6} i}$ $e^( ( pi)/4 i) - e^( ( 7 pi)/6 i)$ using trigonometric functions?

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Answer 1 · 2016-02-19T06:03:06

Pl use Euler's formula to evalute

Explanation:

Euler's formula is $e^{i x} = cos x + i sin x$ $e^(ix) = cosx+isinx$
So $e^{\frac{π}{4} i}$ $e^(pi/4i)$ - $e^{7 \frac{π}{6} i}$ $e^(7pi/6i)$
$= cos (\frac{π}{4}) + i sin (\frac{π}{4}) - (cos (7 \frac{π}{6}) + i sin (7 \frac{π}{6}))$ $= cos(pi/4)+isin(pi/4) - ( cos(7pi/6)+isin(7pi/6))$
$= cos (\frac{π}{4}) - cos (7 \frac{π}{6})) + i (sin (\frac{π}{4} - sin (7 \frac{π}{6}))$ $= cos(pi/4)-cos(7pi/6)) +i ( sin(pi/4-sin(7pi/6))$
$= cos (\frac{π}{4}) - cos (π + \frac{π}{6})) + i (sin (\frac{π}{4} - sin (π + \frac{π}{6}))$ $= cos(pi/4)-cos(pi+pi/6)) +i ( sin(pi/4-sin(pi+pi/6))$
$= cos (\frac{π}{4}) + cos (\frac{π}{6})) + i (sin (\frac{π}{4} + sin (\frac{π}{6}))$ $= cos(pi/4)+cos(pi/6)) +i ( sin(pi/4+sin(pi/6))$
$= \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} + i (\frac{1}{\sqrt{2}} + \frac{1}{2})$ $= 1/sqrt2+sqrt3/2+i(1/sqrt2+1/2)$

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How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{7 π}{6} i}$ $e^( ( pi)/4 i) - e^( ( 7 pi)/6 i)$ using trigonometric functions?

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