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

1 Answer
Cesareo R.
Aug 22, 2016

0.4(cos(40^@)+i sin(40^@))0.4(cos(40∘)+isin(40∘))

Explanation:

Using de Moivre's identity

e^{ix} = cos x + i sin xeix=cosx+isinx

(0.5(cos(100)+isin(100)))(0.8(cos(300)+isin(300))) = (0.5e^{i100^@})(0.8 e^{i 300^@}) = 0.4e^{i 400^@}(0.5(cos(100)+isin(100)))(0.8(cos(300)+isin(300)))=(0.5ei100∘)(0.8ei300∘)=0.4ei400∘

but e^{i 400^@} = e^{i 40^@}e^{i 360^@} = e^{i 40^@}ei400∘=ei40∘ei360∘=ei40∘(periodicity of e^{ix}eix)

Finally

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

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