How do you divide #( 6i-3) / ( 7 i -4 )# in trigonometric form?

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Answer 1 · 2016-02-21T07:28:28

#(3sqrt(13))/13[cos(tan^-1((-1)/18))+i*sin(tan^-1((-1)/18))]# OR

#(3sqrt(13))/13[cos(356.82016988014^@)+i*sin(356.82016988014^@)]" "#

Convert to Trigonometric forms first

#-3+6i=3sqrt5[cos(tan^-1((6)/(-3)))+i sin(tan^-1((6)/(-3)))]#

#-4+7i=sqrt65[cos(tan^-1((7)/(-4)))+i sin(tan^-1((7)/(-4)))]#

Divide equals by equals

#(-3+6i)/(-4+7i)=#

#(sqrt45/sqrt65)[cos(tan^-1((6)/(-3))-tan^-1((7)/(-4)))+i sin(tan^-1((6)/(-3))-tan^-1((7)/(-4)))]#

Take note of the formula:

#tan (A-B)=(Tan A-Tan B)/(1+Tan A* Tan B)#

also

#A-B=Tan^-1 ((Tan A-Tan B)/(1+Tan A* Tan B))#

#(3sqrt(13))/13[cos(tan^-1((-1)/18))+i*sin(tan^-1((-1)/18))]#

#(3sqrt(13))/13[cos(6.2276868019339)+i*sin(6.2276868019339)]" "#radian angles

#(3sqrt(13))/13[cos(356.82016988014^@)+i*sin(356.82016988014^@)]" "#