How do you evaluate $e^{\frac{5 π}{4} i} - e^{\frac{19 π}{12} i}$ $e^( ( 5 pi)/4 i) - e^( ( 19 pi)/12 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{5 π}{4} i} - e^{\frac{19 π}{12} i}$ $e^( ( 5 pi)/4 i) - e^( ( 19 pi)/12 i)$ using trigonometric functions?

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Answer 1 · 2016-06-03T12:30:09

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

Explanation:

$e^{\frac{5 π}{4} i} - e^{\frac{19 π}{12} i}$ $e^( ( 5 pi)/4 i) - e^( ( 19 pi)/12i)$

$= [cos (\frac{5 π}{4}) + i sin (\frac{5 π}{4})] - [cos (\frac{19 π}{12}) + i sin (\frac{19 π}{12})]$ $=[cos((5pi)/4)+isin((5pi)/4)] -[cos((19pi)/12)+isin((19pi)/12)]$

$= [cos (\frac{5 π}{4}) - cos (\frac{19 π}{12})] - i [sin (\frac{19 π}{12}) - sin (\frac{5 π}{4})]$ $=[cos((5pi)/4)-cos((19pi)/12)]-i[sin((19pi)/12)-sin((5pi)/4)]$

$= [cos (\frac{15 π}{12}) - cos (\frac{19 π}{12})] - i [sin (\frac{19 π}{12}) - sin (\frac{15 π}{12})]$ $=[cos((15pi)/12)-cos((19pi)/12)]-i[sin((19pi)/12)-sin((15pi)/12)]$

$= [2 sin (\frac{17 π}{12}) sin (\frac{2 π}{12})] - i [2 cos (\frac{17 π}{12}) sin (\frac{2 π}{12})]$ $=[2sin((17pi)/12)sin((2pi)/12)]-i[2cos((17pi)/12)sin((2pi)/12)]$

$= [2 sin (\frac{17 π}{12}) \times \frac{1}{2}] - i [2 cos (\frac{17 π}{12}) \times \frac{1}{2}]$ $=[2sin((17pi)/12)xx1/2]-i[2cos((17pi)/12)xx1/2]$

$= sin (\frac{17 π}{12}) - i cos (\frac{17 π}{12})$ $=sin((17pi)/12)-icos((17pi)/12)$

Now

$sin (\frac{17 π}{12}) = \sqrt{\frac{1}{2} (1 - cos (\frac{17 π}{6}))} = \sqrt{\frac{1}{2} (1 - cos (3 π - \frac{π}{6}))}$ $sin((17pi)/12)=sqrt(1/2(1-cos((17pi)/6)))=sqrt(1/2(1-cos(3pi-pi/6))$
$= \sqrt{\frac{1}{2} (1 + cos (\frac{π}{6}))} = \sqrt{\frac{1}{2} (1 + cos (\frac{π}{6}))}$ $=sqrt(1/2(1+cos(pi/6)))=sqrt(1/2(1+cos(pi/6)))$

$= \sqrt{\frac{2 + \sqrt{3}}{4}} = \sqrt{\frac{4 + 2 \sqrt{3}}{8}} = \frac{\sqrt{3} + 1}{2 \sqrt{2}}$ $=sqrt((2+sqrt3)/4)=sqrt((4+2sqrt3)/8)=(sqrt3+1)/(2sqrt2)$

and

$cos (\frac{17 π}{12}) = \sqrt{\frac{1}{2} (1 + cos (\frac{17 π}{6}))} = \sqrt{\frac{1}{2} (1 + cos (3 π - \frac{π}{6}))}$ $cos((17pi)/12)=sqrt(1/2(1+cos((17pi)/6)))=sqrt(1/2(1+cos(3pi-pi/6))$

$= \sqrt{\frac{1}{2} (1 - cos (\frac{π}{6}))} = \sqrt{\frac{1}{2} (1 - cos (\frac{π}{6}))}$ $=sqrt(1/2(1-cos(pi/6)))=sqrt(1/2(1-cos(pi/6)))$

$= \sqrt{\frac{2 - \sqrt{3}}{4}} = \sqrt{\frac{4 - 2 \sqrt{3}}{8}} = \frac{\sqrt{3} - 1}{2 \sqrt{2}}$ $=sqrt((2-sqrt3)/4)=sqrt((4-2sqrt3)/8)=(sqrt3-1)/(2sqrt2)$

Hence

$e^{\frac{5 π}{4} i} - e^{\frac{19 π}{12} i}$ $e^( ( 5 pi)/4 i) - e^( ( 19 pi)/12i)$

$= sin (\frac{17 π}{12}) - i cos (\frac{17 π}{12})$ $=sin((17pi)/12)-icos((17pi)/12)$

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

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How do you evaluate $e^{\frac{5 π}{4} i} - e^{\frac{19 π}{12} i}$ $e^( ( 5 pi)/4 i) - e^( ( 19 pi)/12 i)$ using trigonometric functions?

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