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

$\frac{3 \sqrt{13}}{13} \left[\cos \left({\tan}^{-} 1 \left(\frac{- 1}{18}\right)\right) + i \cdot \sin \left({\tan}^{-} 1 \left(\frac{- 1}{18}\right)\right)\right]$ OR

$\frac{3 \sqrt{13}}{13} \left[\cos \left({356.82016988014}^{\circ}\right) + i \cdot \sin \left({356.82016988014}^{\circ}\right)\right] \text{ }$

#### Explanation:

Convert to Trigonometric forms first

$- 3 + 6 i = 3 \sqrt{5} \left[\cos \left({\tan}^{-} 1 \left(\frac{6}{- 3}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{6}{- 3}\right)\right)\right]$

$- 4 + 7 i = \sqrt{65} \left[\cos \left({\tan}^{-} 1 \left(\frac{7}{- 4}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{7}{- 4}\right)\right)\right]$

Divide equals by equals

$\frac{- 3 + 6 i}{- 4 + 7 i} =$

$\left(\frac{\sqrt{45}}{\sqrt{65}}\right) \left[\cos \left({\tan}^{-} 1 \left(\frac{6}{- 3}\right) - {\tan}^{-} 1 \left(\frac{7}{- 4}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{6}{- 3}\right) - {\tan}^{-} 1 \left(\frac{7}{- 4}\right)\right)\right]$

Take note of the formula:

$\tan \left(A - B\right) = \frac{T a n A - T a n B}{1 + T a n A \cdot T a n B}$

also

$A - B = T a {n}^{-} 1 \left(\frac{T a n A - T a n B}{1 + T a n A \cdot T a n B}\right)$

$\frac{3 \sqrt{13}}{13} \left[\cos \left({\tan}^{-} 1 \left(\frac{- 1}{18}\right)\right) + i \cdot \sin \left({\tan}^{-} 1 \left(\frac{- 1}{18}\right)\right)\right]$

$\frac{3 \sqrt{13}}{13} \left[\cos \left(6.2276868019339\right) + i \cdot \sin \left(6.2276868019339\right)\right] \text{ }$radian angles

$\frac{3 \sqrt{13}}{13} \left[\cos \left({356.82016988014}^{\circ}\right) + i \cdot \sin \left({356.82016988014}^{\circ}\right)\right] \text{ }$