How can you use trigonometric functions to simplify # 7 e^( ( 13 pi)/12 i ) # into a non-exponential complex number?

Question

1187 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-25T18:53:09

The answer is #=-7/4((sqrt6+sqrt2))+i7/4((sqrt2-sqrt6))#

We apply Euler's relation

#e^(itheta)=costheta+isintheta#

Therefore,

#7e^(i13/12pi)=7(cos(13/12pi)+isin(13/12pi))#

#cos(13/12pi)=cos(pi+1/12pi)=-cos(1/12pi)#

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

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

#=-(sqrt6+sqrt2)/4#

#sin(13/12pi)=sin(pi+1/12pi)=-sin(1/12pi)#

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

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

#=(sqrt2-sqrt6)/4#

Therefore,

#7e^(i13/12pi)=-7*((sqrt6+sqrt2))/4+i*7((sqrt2-sqrt6))/4#