How do you evaluate # e^( ( 5 pi)/4 i) - e^( ( 3 pi)/8 i)# using trigonometric functions?

1 Answer

Answer:

#2cos((13pi)/16){cos((7pi)/16)+isin((7pi)/16)}#

Explanation:

#e^((5pi)/4i) - e^((3pi)/8i)#

Euler formula : #e^(itheta) = cos(theta) + isin(theta)#

#e^((5pi)/4i) = cos((5pi)/4) + isin((5pi)/4)#
#e^((3pi)/8i) = cos((3pi)/8) + isin((3pi)/8)#

#e^((5pi)/4i)-e^((3pi)/8i)#
#= cos((5pi)/4) + isin((5pi)/4) - (cos((3pi)/8) + isin((3pi)/8))#
#=cos((5pi)/4)-cos((3pi)/8)+i(sin((5pi)/4)-sin((3pi)/8))#

#color(blue)( cos(C) -cos(D) = 2cos((C+D)/2)cos((C-D)/2)#

#color(blue)(sin(C) - sin(D) = 2cos((C+D)/2)sin((C-D)/2)#

#cos((5pi)/4)-cos((3pi)/8) = 2cos(((5pi)/4+(3pi)/8)/2)cos(((5pi)/4-(3pi)/8)/2)#
#cos((5pi)/4)-cos((3pi)/8) = 2cos(1/2((10pi)/8+(3pi)/8))cos((1/2((10pi)/8-(3pi)/8))#
#cos((5pi)/4)-cos((3pi)/8) = 2cos(1/2((13pi)/8)cos((1/2((7pi)/8))#
#color(green)(cos((5pi)/4)-cos((3pi)/8) = 2cos((13pi)/16)cos((7pi)/16)#

#sin((5pi)/4)-sin((3pi)/8) = 2cos(1/2((5pi)/4+(3pi)/8)sin(1/2((5pi)/4-(3pi)/8))#
#sin((5pi)/4)-sin((3pi)/8) = 2cos(1/2((10pi)/8+(3pi)/8)sin(1/2((10pi)/8-(3pi)/8))#
#sin((5pi)/4)-sin((3pi)/8) = 2cos(1/2((13pi)/8)sin(1/2((7pi)/8))#
#color(green)(sin((5pi)/4)-sin((3pi)/8) = 2cos((13pi)/16)sin((7pi)/16)#

#e^((5pi)/4i)-e^((3pi)/8i)#
#=cos((5pi)/4)-cos((3pi)/8)+i(sin((5pi)/4)-sin((3pi)/8))#
#=2cos((13pi)/16)cos((7pi)/16) + i(2cos((13pi)/16)sin((7pi)/16))#
#=2cos((13pi)/16){cos((7pi)/16)+isin((7pi)/16)}#