How do you evaluate #e^( ( 13 pi)/8 i) - e^( ( pi)/4 i)# using trigonometric functions?

1 Answer
Feb 16, 2016

#e^((13pi)/8i) - e^(pi/4i) = (sqrt(2-sqrt(2))-sqrt(2))/2-i*(sqrt(2+sqrt(2))+sqrt(2))/2# ~=
#~= -0.3244-1.6310*i#

Explanation:

First of all, let's recall the Euler's formula that imaginary exponents:
#e^(i*x) = cos(x)+i*sin(x)#

Let's evaluate two terms of the original expression separately and then determine the difference between them.

Based on this formula,
#e^((13pi)/8i) = cos((13pi)/8) + i*sin((13pi)/8)#

Simple property of trigonometric functions that immediately follows from their definition are:
#cos(x+pi) = -cos(x)#
#sin(x+pi) = -sin(x)#
#cos(x+pi/2) = -sin(x)#
#sin(x+pi/2) = cos(x)#

Since #(13pi)/8= pi+(5pi)/8# and #(5pi)/8=pi/2+pi/8#, the expression above can be further evaluated to
#-cos((5pi)/8)-i*sin((5pi)/8) = sin(pi/8)-i*cos(pi/8)#

To determine #sin(pi/8)# and #cos(pi/8)#, recall the formula for #cos# of double angle:
#cos(2x)=2cos^2(x)-1#
From it for #x=pi/8# follows:
#cos^2(pi/8) = 1/2(1+cos(pi/4)) = 1/4(2+sqrt(2))#

#cos(pi/8)=1/2sqrt(2+sqrt(2))#

From #sin^2(x)+cos^2(x)=1# for #x=pi/8# follows
#sin^2(pi/8) = 1-1/4(2+sqrt(2)) = 1/4(2-sqrt(2))#

#sin(pi/8)=1/2sqrt(2-sqrt(2))#

So, the first term in the original expression is
#e^((13pi)/8i) = 1/2sqrt(2-sqrt(2))-i*1/2sqrt(2+sqrt(2))#

The second term in the original expression is
#e^(pi/4i) = cos(pi/4) + i*sin(pi/4) = sqrt(2)/2+i*sqrt(2)/2#

The difference between the first and the second terms is:
#(sqrt(2-sqrt(2))-sqrt(2))/2-i*(sqrt(2+sqrt(2))+sqrt(2))/2#