How do you divide ( 6i-3) / ( 7 i -4 )6i37i4 in trigonometric form?

1 Answer

(3sqrt(13))/13[cos(tan^-1((-1)/18))+i*sin(tan^-1((-1)/18))]31313[cos(tan1(118))+isin(tan1(118))] OR

(3sqrt(13))/13[cos(356.82016988014^@)+i*sin(356.82016988014^@)]" "31313[cos(356.82016988014)+isin(356.82016988014)]

Explanation:

Convert to Trigonometric forms first

-3+6i=3sqrt5[cos(tan^-1((6)/(-3)))+i sin(tan^-1((6)/(-3)))]3+6i=35[cos(tan1(63))+isin(tan1(63))]

-4+7i=sqrt65[cos(tan^-1((7)/(-4)))+i sin(tan^-1((7)/(-4)))]4+7i=65[cos(tan1(74))+isin(tan1(74))]

Divide equals by equals

(-3+6i)/(-4+7i)=3+6i4+7i=

(sqrt45/sqrt65)[cos(tan^-1((6)/(-3))-tan^-1((7)/(-4)))+i sin(tan^-1((6)/(-3))-tan^-1((7)/(-4)))](4565)[cos(tan1(63)tan1(74))+isin(tan1(63)tan1(74))]

Take note of the formula:

tan (A-B)=(Tan A-Tan B)/(1+Tan A* Tan B)tan(AB)=tanAtanB1+tanAtanB

also

A-B=Tan^-1 ((Tan A-Tan B)/(1+Tan A* Tan B))AB=tan1(tanAtanB1+tanAtanB)

(3sqrt(13))/13[cos(tan^-1((-1)/18))+i*sin(tan^-1((-1)/18))]31313[cos(tan1(118))+isin(tan1(118))]

(3sqrt(13))/13[cos(6.2276868019339)+i*sin(6.2276868019339)]" "31313[cos(6.2276868019339)+isin(6.2276868019339)] radian angles

(3sqrt(13))/13[cos(356.82016988014^@)+i*sin(356.82016988014^@)]" "31313[cos(356.82016988014)+isin(356.82016988014)]