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

1 Answer
Dean R.
Apr 23, 2018

#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 #

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