How do you evaluate # e^( ( 3 pi)/8 i) - e^( ( 11 pi)/6 i)# using trigonometric functions?

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Answer 1 · 2018-02-23T17:17:28

#e^((3pi)/8)-e^((11pi)/16)=-2sin((pi)/32)sin((5pi)/32)-2icos((pi)/32)sin((5pi)/32)#

#e^((3pi)/8)-e^((11pi)/16)#
#=cos((3pi)/8)+isin((3pi)/8)-(cos((11pi)/16)+isin((11pi)/16))#

#=(cos((3pi)/8)-cos((11pi)/16))+i(sin((3pi)/8)-sin((11pi)/16))#

#-2sin(((3pi)/8+(11pi)/16)/2)sin(((3pi)/8-(11pi)/16)/2)+i(2cos(((3pi)/8+(11pi)/16)/2)sin(((3pi)/8+(11pi)/16)/2))#

#((3pi)/8+(11pi)/16)/2=pi/2(3/8+11/16)=pi/2((6+11)/16)=17pi/32#

#(17pi)/32=2pi-pi/32#

#((3pi)/8+(11pi)/16)/2=2pi-pi/32#

#((3pi)/8-(11pi)/16)/2=pi/2(3/8-11/16)=pi/2((6-11)/16)=-(5pi)/32#

#(-5pi)/32=2pi-5pi/32#

#((3pi)/8-(11pi)/16)/2=2pi-5pi/32#

#=-2sin(2pi-pi/32)sin(2pi-(5pi)/32)+i(2cos(2pi-pi/32)sin(2pi-(5pi)/32)#

#e^((3pi)/8)-e^((11pi)/16)=-2sin((pi)/32)sin((5pi)/32)-2icos((pi)/32)sin((5pi)/32)#