How can you use trigonometric functions to simplify # 15 e^( ( 17 pi)/12 i ) # into a non-exponential complex number?

Question

1388 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-04T09:15:44

The answer is #=15/4((sqrt2-sqrt6)-i(sqrt2+sqrt6))#

Apply the identity

#z=re^(itheta)=r(costheta+isintheta)#

And

#17/12pi=3/4pi+2/3pi#

Therefore,

#15e^(17/12ipi)=15(cos(3/4pi+2/3pi)+isin(3/4pi+2/3pi))#

#cos(3/4pi+2/3pi)=cos(3/4pi)cos(2/3pi)-sin(3/4pi)sin(2/3pi)#

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

#=sqrt2/4-sqrt6/4#

#sin(3/4pi+2/3pi)=sin(3/4pi)cos(2/3pi)+cos(3/4pi)sin(2/3pi)#

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

#=-sqrt2/4-sqrt6/4#

Finally,

#15e^(17/12ipi)=15/4((sqrt2-sqrt6)+i(-sqrt2-sqrt6))#