Question

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

Answer 1

`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^@))`