Question

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

Answer 1

`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`