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

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Answer 1 · 2016-03-30T15:33:53

$-(sqrt3 -1)/2 - i (sqrt3 +1)/2$

$e^((7pi i)/6)= cos ((7pi)/6) + i sin ((7pi)/6)$

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

Like wise $e^((14pi i)/3)= cos ((14pi)/3) + i sin ((14pi)/3)$

= $cos ((2pi)/3) + i sin ((2pi)/3)$ $[(14pi)/3= 4pi +(2pi)/3= (2pi)/3]$

= $- cos (pi/3) +i sin (pi/3)$

Now combining both expressions, it would be $-sqrt3 /2 -i/2 +1/2 -i sqrt3 /2$

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