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.!