How do you evaluate # e^( ( 5 pi)/4 i) - e^( ( 17 pi)/12 i)# using trigonometric functions?

Question

1679 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-06T15:05:00

The answer is #=(sqrt6-3sqrt2)/4-i(sqrt6-sqrt2)/4#

We need

#cos(a+b)=cosacosb-sinasinb#

#sin(a+b)=sinacosb+sinbcosa#

#sin(pi/4)=cos(pi/4)=sqrt2/2#

#cos(pi/6)=sqrt3/2#

#sin(pi/2)=1/2#

We apply Euler's Formula

#e^(ix)=cosx+isinx#

#e^(5/4pii)=cos(5/4pi)+isin(5/4pi)#

#=-cos(1/4pi)-isin(1/4pi)#

#=-sqrt2/2-isqrt2/2#

#e^(17/12pii)=cos(17/12pi)+isin(17/12pi)#

#=cos(15/12pi+2/12pi)+isin(15/12pi+2/12pi)#

#=cos(5/4pi)cos(1/6pi)-sin(5/4pi)sin(1/6pi)+i(sin(5/4pi)cos(1/6pi)+cos(5/4pi)sin(1/6pi))#

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

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

Therefore,

#e^(5/4pii)-e^(17/12pii)=-sqrt2/2-isqrt2/2-((sqrt2-sqrt6)/4+i(-sqrt6-sqrt2)/4)#

#=(sqrt6-3sqrt2)/4-i(sqrt6-sqrt2)/4#