Question

How do you evaluate `e^( ( 7 pi)/4 i) - e^( ( 13 pi)/12 i)` using trigonometric functions?

Answer 1

`\frac{sqrt(3)+3}{2sqrt(2)}-i\frac{sqrt(3)+3}{2sqrt(2)}`

Explanation:

Since `e^{itheta} = cos(theta)+isin(theta)`, your expression becomes

`cos((7pi)/4)+isin((7pi)/4)-(cos((13pi)/12)+isin((13pi)/12))`

Now let's work with the angles: since `(7pi)/4 = 2pi-pi/4 = -pi/4`, we have that

• `cos((7pi)/4) = cos(-pi/4) = cos(pi/4) = sqrt(2)/2`
• `sin((7pi)/4) = sin(-pi/4) = -sin(pi/4) = -sqrt(2)/2`

On the other hand, we have that `(13pi)/12 = pi+pi/12`, and so the transformations are

• `cos((13pi)/12) = cos(pi+pi/12) = -cos(pi/12) = -\frac{sqrt(3)+1}{2sqrt(2)}`
• `sin((13pi)/12) = sin(pi+pi/12) = -sin(pi/12) = -\frac{sqrt(3)-1}{2sqrt(2)}`

So, your expression becomes

`sqrt(2)/2-isqrt(2)/2-(-\frac{sqrt(3)+1}{2sqrt(2)}-i(-\frac{sqrt(3)-1}{2sqrt(2)}))`

Which simplifies into

`sqrt(2)/2-isqrt(2)/2+\frac{sqrt(3)+1}{2sqrt(2)}-i\frac{sqrt(3)-1}{2sqrt(2)}`

You can write it with just one denominator:

`\frac{sqrt(3)+3}{2sqrt(2)}-i\frac{sqrt(3)+3}{2sqrt(2)}`