Question

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

Answer 1

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

Explanation:

`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`