How can you use trigonometric functions to simplify # 9 e^( ( 17 pi)/12 i ) # into a non-exponential complex number?

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Answer 1 · 2018-07-26T16:50:18

The answer is #=9/4((-sqrt6+sqrt2)-i(sqrt2+sqrt6))#

The Euler's Identity is

#e^(itheta)=costheta+isintheta#

Therefore,

#z=9e^(17/12pii)#

#=9cos(17/12pi)+isin(17/12pi)#

#cos(17/12pi)=cos(7/6pi+1/4pi)#

#=cos(7/6pi)cos(1/4pi)-sin(7/6pi)sin(1/4pi)#

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

#=1/4(-sqrt6+sqrt2)#

#sin(17/12pi)=sin(7/6pi+1/4pi)#

#=sin(7/6pi)cos(1/4pi)+cos(7/6pi)sin(1/4pi)#

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

#=-1/4(sqrt2+sqrt6)#

Finally,

#z=9/4((-sqrt6+sqrt2)-i(sqrt2+sqrt6))#