Question

How do you evaluate `e^( ( 23 pi)/12 i) - e^( ( 13 pi)/12 i)` using trigonometric functions?

Answer 1

`e^{i ({ 23 pi}/12) } - e^{i ({13 \pi}/12) } = {sqrt{6} + sqrt{2}}/2`

Explanation:

`e^{i ({ 23 pi}/12) } - e^{i ({13 \pi}/12) }`

`= e^{-2 pi i} e^{i ({ 23 pi}/12) } - e^{i pi} e^{i ({\pi}/12) }`

`= e^{i (-{ pi}/12) } + e^{i ({\pi}/12) }`

`= cos(pi/12) - i \sin (pi/12) + cos(\pi/12) + i \sin(pi/12)`

`= 2 \cos(pi/12)`

`pi/12` is `15^circ`. We avoid the nested radical with the difference formula:

`cos(pi/12) = cos(45^circ - 30^circ) = cos 45^circ cos 30^circ + sin 45^circ sin 30^circ` `= \sqrt{2}/2 ( \sqrt{3}/2 + 1/2) = {sqrt{6} + sqrt{2}}/4`

`e^{i ({ 23 pi}/12) } - e^{i ({13 \pi}/12) } = {sqrt{6} + sqrt{2}}/2`