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

1 Answer
Jan 6, 2016

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)}