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

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Answer 1 · 2016-01-31T03:28:28

$e^{i (7pi)/4} - e^{i pi/6} = (sqrt2-sqrt3)/2 - i ((1+sqrt2)/2)$

Use the Euler's Formula, which states that $e^{i theta} -= cos(theta) + i sin(theta)$ . (Proof omitted)

Therefore,

$e^{i (7pi)/4} - e^{i pi/6} = (cos((7pi)/4) + i sin((7pi)/4)) - (cos(pi/6) + i sin(pi/6))$

$= (sqrt2/2 + i (-sqrt2/2)) - (sqrt3/2 + i (1/2))$

$= (sqrt2-sqrt3)/2 - i ((1+sqrt2)/2)$

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