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

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Answer 1 · 2016-08-22T13:42:33

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

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

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

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)))?