How do you divide $\frac{i + 12}{9 i - 10}$ $(i+12) / (9i-10)$ in trigonometric form?

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How do you divide $\frac{i + 12}{9 i - 10}$ $(i+12) / (9i-10)$ in trigonometric form?

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Answer 1 · 2017-04-03T17:08:19

Use $\frac{r_{1} (cos (θ_{1}) + i sin (θ_{1}))}{r_{2} (cos (θ_{2}) + i sin (θ_{2}))} = \frac{r_{1}}{r_{2}} (cos (θ_{1} - θ_{2}) + i sin (θ_{1} - θ_{2}))$ $(r_1(cos(theta_1)+isin(theta_1)))/(r_2(cos(theta_2)+isin(theta_2))) = r_1/r_2(cos(theta_1-theta_2)+isin(theta_1-theta_2))$

Explanation:

Compute $r_{1}$ $r_1$ :

$r_{1} = \sqrt{12^{2} + 1^{2}}$ $r_1 = sqrt(12^2+1^2)$

$r_{1} = \sqrt{145}$ $r_1 = sqrt(145)$

Compute $θ_{1}$ $theta_1$

$θ_{1} = {tan}^{- 1} (\frac{1}{12})$ $theta_1 = tan^-1(1/12)$

Compute $r_{2}$ $r_2$ :

$r_{2} = \sqrt{- 10^{2} + 9^{2}}$ $r_2 = sqrt(-10^2+9^2)$

$r_{2} = \sqrt{181}$ $r_2 = sqrt(181)$

Compute $θ_{2}$ $theta_2$ (second quadrant)

$θ_{2} = {tan}^{- 1} (\frac{9}{- 10}) + π$ $theta_2 = tan^-1(9/-10)+pi$

The resulting division is:

$\sqrt{\frac{145}{181}} (cos ({tan}^{- 1} (\frac{1}{12}) - {tan}^{- 1} (\frac{9}{- 10}) - π) + i sin ({tan}^{- 1} (\frac{1}{12}) - {tan}^{- 1} (\frac{9}{- 10}) - π))$ $sqrt(145/181)(cos(tan^-1(1/12)-tan^-1(9/-10)-pi)+isin(tan^-1(1/12)-tan^-1(9/-10)-pi))$

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How do you divide $\frac{i + 12}{9 i - 10}$ $(i+12) / (9i-10)$ in trigonometric form?

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Explanation:

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How do you divide $\frac{i + 12}{9 i - 10}$ $(i+12) / (9i-10)$ in trigonometric form?