How do you evaluate # e^( ( 13 pi)/8 i) - e^( (5 pi)/12 i)# using trigonometric functions?

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Answer 1 · 2018-01-25T10:27:48

#e^((13pi)/8i)-e^((5pi)/12i)~~0.12-1.89i#

We can represent #ae^(ix)# in trig form as #ae^(ix)=a(cosx+isinx)#

Using this for #e^((13pi)/8i)-e^((5pi)/12i)# gives us:
#(cos((13pi)/8)+isin((13pi)/8))-(cos((5pi)/12)+isin((5pi)/12))#

#=cos((13pi)/8)+isin((13pi)/8)-cos((5pi)/12)-isin((5pi)/12))#

#=cos((13pi)/8)-cos((5pi)/12)-isin((5pi)/12)+isin((13pi)/8)#

#=(cos((13pi)/8)-cos((5pi)/12))-i(sin((5pi)/12)-sin((13pi)/8))#

#~~0.1238643873-i(1.889805359)#

#=0.1238643873-1.889805359i#

#~~0.12-1.89i#