How do you evaluate #e^( ( 13 pi)/8 i) - e^( ( 3 pi)/4 i)# using trigonometric functions?

Question

1109 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-07T09:33:25

#=1/sqrt2(1-sqrt(1-1/sqrt2)+i(sqrt(1+1/sqrt2) -1))#
= #0.3242+i0.2168#, nearly. After some time, I would edit my answer myself, for bugs, if any.

Use

#e^(i2pi)=cos2pi+isin2pi=1# and

#e^(ipi)=cospi+isinpi=-1#

The given expression is

#e^(ipi)e^(i5/8pi)-e^(ipi)e^(-i1/4pi)#

#=(cos(112.5^o)+isin(112.5^o))+(cos45^0-isin45^o))#

#=-sin22.5^o +cos45^o+i(cos22.5^o -sin45^o)#

#=1/sqrt2-sqrt((1-cos45^0)/2)+i(-1/sqrt2+sqrt((1+cos45^0)/2))#

#=1/sqrt2(1-sqrt(1-1/sqrt2)+i(sqrt(1+1/sqrt2) -1))#

#=0.3244+i0.2168#, nearly.