How do you evaluate # e^( ( 7 pi)/4 i) - e^( ( 5 pi)/6 i)# using trigonometric functions?

Question

1198 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-17T18:32:03

=- 2sin15(cos15+isin15)

=# cos ((7pi)/4) +i sin ((7pi)/4) - cos( (5pi)/6) - i sin ((7pi)/6)#

= #cos ((7pi)/4) -cos ((5pi)/6) +i [ sin((7pi)/4) - sin((5pi)/6)]#

= #-2 sin((31pi)/12) sin ((11pi)/12) +2i sin((11pi)/12) cos ((31pi)/12)#

=#2sin ((11pi)/12)[ -sin((31pi)/12) +i cos((31pi)/12)]#

=#2sin ((11pi)/12)[ -sin((7pi)/12) +i cos((7pi)/12)]#

=2 sin (pi/12)[ -sin105+ i cos 105]

= 2 sin 15(-cos 15 - i sin 15)

=- 2sin15(cos15+isin15)