"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
