How can you use trigonometric functions to simplify # 8 e^( ( 19 pi)/12 i ) # into a non-exponential complex number?

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Answer 1 · 2017-12-04T12:54:29

The answer is #=2(sqrt6-sqrt2)-2i(sqrt6+sqrt2)#

Apply Euler's identity

#e^(itheta)=costheta+ isin theta#

#19/12pi=5/4pi+1/3pi#

#cos(19/12pi)=cos(5/4pi+1/3pi)#

#=cos(5/4pi)cos(1/3pi)-sin(5/4pi)sin(1/3pi)#

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

#=(sqrt6-sqrt2)/4#

#sin(19/12pi)=sin(5/4pi+1/3pi)#

#=sin(5/4pi)cos(1/3pi)+sin(1/3pi)cos(5/4pi)#

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

#=-(sqrt2+sqrt6)/4#

Therefore,

#8e^(19/12pi)=8cos(19/12pi)+i8sin(19/12pi)#

#=8*((sqrt6-sqrt2)/4)-8i(-(sqrt2+sqrt6)/4)#

#=2(sqrt6-sqrt2)-2i(sqrt6+sqrt2)#