How do you perform the operation in trigonometric form $(6(cos40+isin40))/(7(cos100+isin100))$ ?

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Answer 1 · 2016-09-17T04:29:30

$3/7(1-isqrt3)$ .

Let $z_1=6(cos40+isin40), and, z_2=7(cos100+isin100)$

Knowing that, $costheta+isintheta=e^(itheta)$ , we have,

$z_1=r_1e^(itheta_1), and, z_2=r_2e^(itheta_2)$ , then,

$r_1=6, theta_1=40, r_2=7, theta_2=100$ .

$:." The Expression="z_1/z_2=(r_1e^(itheta_1))/(r_2e^(itheta_2))$

$=(r_1/r_2)*e^(itheta_1-itheta_2)$

$=(r_1/r_2)*e^(i(theta_1-theta_2))$

$=(6/7)*e^(i(40-100))$

$=(6/7)*e^(i(-60))$

$=(6/7)(cos(-60)+isin(-60))$

$=(6/7)(cos60-isin60)$

$=(6/7)(1/2-isqrt3/2)$

$=3/7(1-isqrt3)$ .

Enjoy maths.!

How do you perform the operation in trigonometric form (6(cos40+isin40))/(7(cos100+isin100))(6(cos40+isin40))/(7(cos100+isin100))?