How do you evaluate $e^{\frac{11 π}{6} i} - e^{\frac{17 π}{12} i}$ $e^( ( 11 pi)/6 i) - e^( ( 17 pi)/12 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{11 π}{6} i} - e^{\frac{17 π}{12} i}$ $e^( ( 11 pi)/6 i) - e^( ( 17 pi)/12 i)$ using trigonometric functions?

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Answer 1 · 2016-10-30T21:02:45

$e^{11 \frac{π}{6} i} - e^{\frac{17 π}{12} i} = \frac{2 \sqrt{3} + \sqrt{6} - \sqrt{2}}{4} + i \frac{\sqrt{2} + \sqrt{6} - 2}{4}$ $e^(11pi/6i)-e^((17pi)/12i)=(2sqrt3+sqrt6-sqrt2)/4+i(sqrt2+sqrt6-2)/4$

Explanation:

Evaluating the expression is determined by using Euler's theorem

$e^{i x} = cos x + i sin x$ $color(blue)(e^(ix)=cosx+isinx)$

$e^{11 \frac{π}{6} i} = cos (11 \frac{π}{6}) + i sin (11 \frac{π}{6})$ $e^(11pi/6i)=color(blue)(cos(11pi/6)+isin(11pi/6)$
$e^{11 \frac{π}{6} i} = cos (\frac{12 π}{6} - \frac{π}{6}) + i sin (\frac{12 π}{6} - \frac{π}{6})$ $e^(11pi/6i)=cos((12pi)/6-pi/6)+isin((12pi)/6-pi/6)$
$e^{11 \frac{π}{6} i} = cos (2 π - \frac{π}{6}) + i sin (2 π - \frac{π}{6})$ $e^(11pi/6i)=cos(2pi-pi/6)+isin(2pi-pi/6)$
$e^{11 \frac{π}{6} i} = cos (- \frac{π}{6}) + i sin (- \frac{π}{6})$ $e^(11pi/6i)=cos(-pi/6)+isin(-pi/6)$

Applying the trigonometric identities:

$cos (- α) = cos α$ $color(green)(cos(-alpha)=cosalpha)$
$sin (- α) = - sin α$ $color(green)(sin(-alpha)=-sinalpha)$

Let us continue computing $e^{\frac{11 π}{6}}$ $e^((11pi)/6)$

$e^{11 \frac{π}{6} i} = cos (- \frac{π}{6}) + i sin (- \frac{π}{6})$ $e^(11pi/6i)=cos(-pi/6)+isin(-pi/6)$
$e^{11 \frac{π}{6} i} = cos (\frac{π}{6}) - i sin (\frac{π}{6})$ $e^(11pi/6i)=color(green)(cos(pi/6))color(green)(-isin(pi/6))$
$e^{11 \frac{π}{6} i} = \frac{\sqrt{3}}{2} - i \frac{1}{2}$ $e^(11pi/6i)=sqrt3/2-i1/2$

Let us compute $e^{\frac{17 π}{12} i}$ $e^((17pi)/12i)$

$e^{\frac{17 π}{12} i} = cos (\frac{17 π}{12}) + i sin (\frac{17 π}{12})$ $e^((17pi)/12i)=color(blue)(cos((17pi)/12)+isin((17pi)/12)$
$e^{\frac{17 π}{12} i} = cos (\frac{12 π}{12} + \frac{5 π}{12}) + i sin (\frac{12 π}{12} + \frac{5 π}{12})$ $e^((17pi)/12i)=cos((12pi)/12+(5pi)/12)+isin((12pi)/12+(5pi)/12)$
$e^{\frac{17 π}{12} i} = cos (π + \frac{5 π}{12}) + i sin (π + \frac{5 π}{12})$ $e^((17pi)/12i)=cos(pi+(5pi)/12)+isin(pi+(5pi)/12)$

Applying the trigonometric identities

$cos (π + α) = - cos α$ $color(brown)(cos(pi+alpha)=-cosalpha)$
$sin (π + α) = - sin α$ $color(brown)(sin(pi+alpha)=-sinalpha)$

Let us continue computing $e^{\frac{17 π}{12} i}$ $e^((17pi)/12i)$
$e^{\frac{17 π}{12} i} = cos (π + \frac{5 π}{12}) + i sin (π + \frac{5 π}{12})$ $e^((17pi)/12i)=cos(pi+(5pi)/12)+isin(pi+(5pi)/12)$
$e^{\frac{17 π}{12} i} = - cos (\frac{5 π}{12}) - i sin (\frac{5 π}{12})$ $e^((17pi)/12i)=color(brown)(-cos((5pi)/12)color(brown)-isin((5pi)/12)$
$e^{\frac{17 π}{12} i} = - \frac{\sqrt{6} - \sqrt{2}}{4} - i \frac{\sqrt{2} + \sqrt{6}}{4}$ $e^((17pi)/12i)=-(sqrt6-sqrt2)/4-i(sqrt2+sqrt6)/4$
$e^{\frac{17 π}{12} i} = \frac{- \sqrt{6} + \sqrt{2}}{4} - i \frac{\sqrt{2} + \sqrt{6}}{4}$ $e^((17pi)/12i)=(-sqrt6+sqrt2)/4-i(sqrt2+sqrt6)/4$

hint:To evaluate $cos (\frac{5 π}{12}) and sin (\frac{5 π}{12})$ $cos((5pi)/12) and sin((5pi)/12)$
$cos (\frac{5 π}{12}) = cos (\frac{π}{6} + \frac{π}{4})$ $cos((5pi)/12)=cos(pi/6+pi/4)$
$sin (\frac{5 π}{12}) = sin (\frac{π}{6} + \frac{π}{4})$ $sin((5pi)/12)=sin(pi/6+pi/4)$

So,

$e^{11 \frac{π}{6} i} - e^{\frac{17 π}{12} i}$ $color(blue)(e^(11pi/6i)-e^((17pi)/12i)$

$= \frac{\sqrt{3}}{2} - i \frac{1}{2} - (\frac{- \sqrt{6} + \sqrt{2}}{4} - i \frac{\sqrt{2} + \sqrt{6}}{4})$ $=sqrt3/2-i1/2-((-sqrt6+sqrt2)/4-i(sqrt2+sqrt6)/4)$

$= \frac{\sqrt{3}}{2} - i \frac{1}{2} - (\frac{- \sqrt{6} + \sqrt{2}}{4}) + i \frac{\sqrt{2} + \sqrt{6}}{4}$ $=sqrt3/2-i1/2-((-sqrt6+sqrt2)/4)+i(sqrt2+sqrt6)/4$

$= \frac{\sqrt{3}}{2} - i \frac{1}{2} + \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} + i \frac{\sqrt{2} + \sqrt{6}}{4}$ $=sqrt3/2-i1/2+sqrt6/4-sqrt2/4+i(sqrt2+sqrt6)/4$

$= \frac{2 \sqrt{3}}{4} - \frac{2 i}{4} + \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} + i \frac{\sqrt{2} + \sqrt{6}}{4}$ $=(2sqrt3)/4-(2i)/4+sqrt6/4-sqrt2/4+i(sqrt2+sqrt6)/4$

$= \frac{2 \sqrt{3}}{4} + \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} + i \frac{\sqrt{2} + \sqrt{6}}{4} - \frac{2 i}{4}$ $=(2sqrt3)/4+sqrt6/4-sqrt2/4+i(sqrt2+sqrt6)/4-(2i)/4$

$= \frac{2 \sqrt{3} + \sqrt{6} - \sqrt{2}}{4} + i \frac{\sqrt{2} + \sqrt{6} - 2}{4}$ $=(2sqrt3+sqrt6-sqrt2)/4+i(sqrt2+sqrt6-2)/4$

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How do you evaluate $e^{\frac{11 π}{6} i} - e^{\frac{17 π}{12} i}$ $e^( ( 11 pi)/6 i) - e^( ( 17 pi)/12 i)$ using trigonometric functions?

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