Question

How do you evaluate `e^( ( 7 pi)/6 i) - e^( ( 4 pi)/3 i)` using trigonometric functions?

Answer 1

`2[sin ((5pi)/4)sin (pi/12) + i cos ((5pi)/4) sin (pi/12)]`

Explanation:

`e^((7pi i)/6 )= cos ((7pi)/6) - i sin ((7pi)/6)`

`e^((4pi i)/3)= cos ((4pi)/3) -i sin ((4pi)/3)`

The required simplification would be

`cos ((7pi )/6) - cos((4pi )/3) +i[sin ((4pi)/3) - sin ((7pi)/6)]`

= `2 sin( ((7pi)/6+ (4pi)/3)/2) sin( ((4pi)/3 -(7pi)/6)/2) +i[2 cos(((4pi)/3 +(7pi)/6)/2) sin (((4pi)/3 -(7pi)/6)/2)]`

= `2[sin ((5pi)/4)sin (pi/12) + i cos ((5pi)/4) sin (pi/12)]`