# "Recall the definition of the complex exponential:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad e^{ i \theta } \ = \ cos(\theta) + i sin(\theta). #
# "Using this with the given expression, we have:" #
# \qquad \qquad \qquad e^{ {13\pi}/4 i } - e^{ {7\pi}/6 i } \ = #
# [ cos( {13\pi}/4 ) + sin( {13\pi}/4 ) i ] - [ cos( {7\pi}/6 ) + sin( {7\pi}/6 ) i ] \ =#
# [ cos( {13\pi}/4 ) + cos( {7\pi}/6 ) ] - [ sin( {13\pi}/4 )+ sin( {7\pi}/6 ) ] i. #
# :. \qquad \ \ e^{ {13\pi}/4 i } - e^{ {7\pi}/6 i } \ = #
# [ cos( {13\pi}/4 ) + cos( {7\pi}/6 ) ] - [ sin( {13\pi}/4 )+ sin( {7\pi}/6 ) ] i. \quad \ (1) #
# "Now we must calculate the trig values in the above:" #
# \qquad \ cos( {13\pi}/4 ), \qquad sin( {13\pi}/4 ); \qquad cos( {7\pi}/6 ), qquad sin( {7\pi}/6 ). \qquad \ (2) #
# "For the first two trig values we note:" #
# cos( {13\pi}/4 ) \ = \ cos( 13/4 \pi ) \ = \ cos( [3 1/4] \pi ) #
# \qquad \qquad \qquad \qquad \qquad = \ cos( [ 2 + 1 1/4] \pi ) \ = \ cos( 2 \pi + [1 1/4] \pi ) #
# \qquad \qquad \qquad \qquad \qquad = \ cos( [ 1 1/4] \pi ) \ = \ cos( \pi + 1/4 \pi ) #
# \qquad \qquad \qquad \qquad \qquad = \ - cos( 1/4 \pi ), \qquad \qquad "as" \quad \pi + 1/4 \pi \ \in \ "Quadrant III" #
# \qquad \qquad \qquad \qquad \qquad = \ - cos( 45^@ ) #
# \qquad \qquad = \ - \sqrt{2} / 2, \qquad \quad "remembering the 45-45-90 right triangle." #
# "Similarly, proceeding with" \quad sin( {13\pi}/4 ), "we summarize:" #
# sin( {13\pi}/4 ) \ = \ sin( \pi + 1/4 \pi ) #
# \qquad \qquad \qquad \qquad \qquad = \ - sin( 1/4 \pi ), \qquad \qquad "as" \quad \pi + 1/4 \pi \ \in \ "Quadrant III" #
# \qquad \qquad \qquad \qquad \qquad = \ - sin( 45^@ ) #
# \qquad \qquad = \ - \sqrt{2} / 2, \qquad \quad "remembering the 45-45-90 right triangle." #
# "So we have:" #
# \qquad \qquad \quad cos( {13\pi}/4 ) \ = \ - \sqrt{2} / 2, \qquad \qquad sin( {13\pi}/4 ) \ = \ - \sqrt{2} / 2. \quad (3) #
# "Now, for the last two trig values in line (2) we note:" #
# cos( {7\pi}/6 ) \ = \ cos( 7/6 \pi ) \ = \ cos( [1 1/6] \pi ) #
# \qquad \qquad \qquad \qquad \qquad = \ cos( [ 1 + 1/6 ] \pi ) \ = \ cos( \pi + 1/6 \pi ) #
# \qquad \qquad \qquad \qquad \qquad = \ - cos( 1/6 \pi ), \qquad \qquad "as" \quad \pi + 1/6 \pi \ \in \ "Quadrant III" #
# \qquad \qquad \qquad \qquad \qquad = \ - cos( 30^@ ) #
# \qquad \qquad = \ - \sqrt{3} / 2, \qquad \quad "remembering the 30-60-90 right triangle." #
# "Similarly, proceeding with" \quad sin( 1/6 \pi ), "we summarize:" #
# sin( 7/6 \pi ) \ = \ sin( \pi + 1/6 \pi ) #
# \qquad \qquad \qquad \qquad \qquad = \ - sin( 1/6 \pi ), \qquad \qquad "as" \quad \pi + 1/6 \pi \ \in \ "Quadrant III" #
# \qquad \qquad \qquad \qquad \qquad = \ - sin( 30^@ ) #
# \qquad \qquad = \ - 1 / 2, \qquad \quad \ \ \ "remembering the 30-60-90 right triangle." #
# "So we have:" #
# \qquad \qquad \qquad \quad cos( {7\pi}/6 ) \ = \ - \sqrt{3} / 2, \qquad \qquad sin( {7\pi}/6 ) \ = \ - 1 / 2. \qquad \quad (4) #
# "So, substituting the trig results we have in eqns. (3) & (4), into" #
# "our original eqn. (1), we have now:" #
# \qquad \qquad \qquad e^{ {13\pi}/4 i } - e^{ {7\pi}/6 i } \ = #
# [ cos( {13\pi}/4 ) + cos( {7\pi}/6 ) ] - [ sin( {13\pi}/4 )+ sin( {7\pi}/6 ) ] i \ = #
# \qquad \qquad \qquad \qquad [ - sqrt{2} / 2 - \sqrt{3} / 2 ] - [ - \sqrt{2} / 2 - 1 / 2 ] i \ = #
# \qquad \qquad \qquad \qquad \qquad - 1/2 ( [ sqrt{2} + \sqrt{3} ] - [ \sqrt{2} + 1 ] i ). #
# "This is our answer." #
# "So, summarizing:" #
# \qquad \qquad \qquad \ \ e^{ {13\pi}/4 i } - e^{ {7\pi}/6 i } \ = \ - 1/2 ( [ sqrt{2} + \sqrt{3} ] - [ \sqrt{2} + 1 ] i ). \qquad \ \ square #
