How can you use trigonometric functions to simplify # 5 e^( ( 13 pi)/12 i ) # into a non-exponential complex number?

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Answer 1 · 2017-08-11T09:09:51

The answer is #=-5/4(sqrt6+sqrt2)+i5/4(sqrt2-sqrt6)#

We apply Euler's relation

#e^(itheta)=costheta+isintheta#

#5e^(13/12ipi)=5(cos(13/12pi)+isin(13/12pi))#

#cos(13/12pi)=cos(1/3pi+3/4pi)#

#=cos(1/3pi)cos(3/4pi)-sin(1/3pi)sin(3/4pi)#

#=-1/2*sqrt2/2-sqrt3/2*sqrt2/2#

#=-(sqrt6+sqrt2)/4#

#sin(13/12pi)=sin(1/3pi+3/4pi)#

#=sin(1/3pi)cos(3/4pi)+cos(1/3pi)sin(3/4pi)#

#=-sqrt3/2*sqrt2/2+1/2*sqrt2/2#

#=(sqrt2-sqrt6)/4#

Therefore,

#5e^(13/12ipi)=-5/4(sqrt6+sqrt2)+i5/4(sqrt2-sqrt6)#