How do you perform the operation in trigonometric form #(0.5(cos(100)+isin(100)))(0.8(cos(300)+isin(300)))#?

1 Answer
Aug 22, 2016

#0.4(cos(40^@)+i sin(40^@))#

Explanation:

Using de Moivre's identity

#e^{ix} = cos x + i sin x#

#(0.5(cos(100)+isin(100)))(0.8(cos(300)+isin(300))) = (0.5e^{i100^@})(0.8 e^{i 300^@}) = 0.4e^{i 400^@}#

but #e^{i 400^@} = e^{i 40^@}e^{i 360^@} = e^{i 40^@}#(periodicity of #e^{ix}#)

Finally

#(0.5(cos(100^@)+isin(100^@)))(0.8(cos(300^@)+isin(300^@))) = 0.4(cos(40^@)+i sin(40^@))#