How do you evaluate # e^( ( 7 pi)/6 i) - e^( ( 7 pi)/8 i)# using trigonometric functions?

1 Answer
Ratnaker Mehta
Jun 14, 2016

#{(sqrt(2+sqrt2)-sqrt3)-i(sqrt(2-sqrt2)+1)}/2.#

Explanation:

We have, #e^(itheta)=costheta+isintheta.#

#:.e^((7pi/6)i)-e^((7pi/8)i)=cos7pi/6+isin7pi/6-cos7pi/8-isin7pi/8=cos(pi+pi/6)+isin(pi+pi/6)-cos(pi-pi/8)-isin(pi-pi/8)=-cos(pi/6)-isin(pi/6)+cos(pi/8)-isin(pi/8)=-sqrt3/2-i/2+(sqrt(2+sqrt2))/2-i(sqrt(2-sqrt2))/2=(sqrt(2+sqrt2))/2-sqrt3/2-i(sqrt(2-sqrt2))/2-i/2={(sqrt(2+sqrt2)-sqrt3)-i(sqrt(2-sqrt2)+1)}/2.#

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