How do you evaluate e^( ( 7 pi)/4 i) - e^( ( 13 pi)/12 i)e7π4i−e13π12i using trigonometric functions?

1 Answer
KillerBunny
Dec 3, 2016

\frac{sqrt(3)+3}{2sqrt(2)}-i\frac{sqrt(3)+3}{2sqrt(2)}√3+32√2−i√3+32√2

Explanation:

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

cos((7pi)/4)+isin((7pi)/4)-(cos((13pi)/12)+isin((13pi)/12))cos(7π4)+isin(7π4)−(cos(13π12)+isin(13π12))

Now let's work with the angles: since (7pi)/4 = 2pi-pi/4 = -pi/47π4=2π−π4=−π4, we have that

  • cos((7pi)/4) = cos(-pi/4) = cos(pi/4) = sqrt(2)/2cos(7π4)=cos(−π4)=cos(π4)=√22
  • sin((7pi)/4) = sin(-pi/4) = -sin(pi/4) = -sqrt(2)/2sin(7π4)=sin(−π4)=−sin(π4)=−√22

On the other hand, we have that (13pi)/12 = pi+pi/1213π12=π+π12, and so the transformations are

  • cos((13pi)/12) = cos(pi+pi/12) = -cos(pi/12) = -\frac{sqrt(3)+1}{2sqrt(2)}cos(13π12)=cos(π+π12)=−cos(π12)=−√3+12√2
  • sin((13pi)/12) = sin(pi+pi/12) = -sin(pi/12) = -\frac{sqrt(3)-1}{2sqrt(2)}sin(13π12)=sin(π+π12)=−sin(π12)=−√3−12√2

So, your expression becomes

sqrt(2)/2-isqrt(2)/2-(-\frac{sqrt(3)+1}{2sqrt(2)}-i(-\frac{sqrt(3)-1}{2sqrt(2)}))√22−i√22−(−√3+12√2−i(−√3−12√2))

Which simplifies into

sqrt(2)/2-isqrt(2)/2+\frac{sqrt(3)+1}{2sqrt(2)}-i\frac{sqrt(3)-1}{2sqrt(2)}√22−i√22+√3+12√2−i√3−12√2

You can write it with just one denominator:

\frac{sqrt(3)+3}{2sqrt(2)}-i\frac{sqrt(3)+3}{2sqrt(2)}√3+32√2−i√3+32√2

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