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

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

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