How do you evaluate # e^( ( 13 pi)/8 i) - e^( (5 pi)/6 i)# using trigonometric functions?

Pet peeve: every trig problem is 30,60,90 or 45.45.90. This one's both.

# e^{i ({13 pi}/8) } - e^{i ({5 pi}/6)} #

# = e^{i ({13 pi}/8 ) } - e^{i ({5 pi}/6)} #

#= cos({13 \pi}/8) + i \sin ({13 \pi}/8) - (cos({5\pi}/6) +i \sin ({5pi}/6)) #

# = cos(2pi - {13 pi}/ 8) - i sin(2pi - {13pi}/8) - (- cos(pi-{5 pi}/6) + i sin(pi -{5pi}/6)) #

# = cos({3 pi}/ 8) - i sin({3pi}/8) + cos(pi/6) + i sin(pi/6) #

We've evaluated the expression using trigonometric functions. We can go further and evaluate those trigonometric functions.

Let's start by noting #{3 pi}/4# is #135^circ# so #{3pi}/8# is #135^circ / 2 = 67.5^circ#

That's in the first quadrant, so we can use the half angle formula with a + sign:

#cos 67.5^circ = + \sqrt{ 1/2 ( 1 + cos 135^circ) } = sqrt{1/2( 1 - sqrt{2}/2)}#

#sin 67.5^circ = + \sqrt{ 1/2 ( 1 - cos 135^circ) } = sqrt{1/2( 1 + sqrt{2}/2)}#

# = (sqrt{1/2( 1 - sqrt{2}/2)} + sqrt{3}/2 ) + i(1/2 - sqrt{1/2( 1 + sqrt{2}/2)} ) #

# = 1/2 ( sqrt{2 - sqrt{2)} + sqrt{3}) + i/2(1-sqrt{2 + sqrt{2}} ) #

That may be right. I'll leave it here.