How do you divide $\frac{9 i - 2}{- 5 i + 2}$ $( 9i- 2) / (- 5 i+ 2 )$ in trigonometric form?

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How do you divide $\frac{9 i - 2}{- 5 i + 2}$ $( 9i- 2) / (- 5 i+ 2 )$ in trigonometric form?

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Answer 1 · 2016-12-20T17:32:02

$= \frac{1}{29} (41 + 8 i)$ $=1/29 (41+8i)$

Explanation:

Rewrite 9i-2 as -2+9i. Now square and the numbers -2 and 9, which would be 85. Its square root is $\sqrt{85}$ $sqrt85$ . Multiply and divide the expression -2 +9i with $\sqrt{85}$ $sqrt85$ as follows:

$\sqrt{85} (- \frac{2}{\sqrt{85}} + i \frac{9}{\sqrt{85}})$ $sqrt85 (-2/sqrt85 +i 9/sqrt85)$

If $θ$ $theta$ is some angle then let $cos θ = - \frac{2}{\sqrt{85}} and sin θ = \frac{9}{\sqrt{85}}$ $cos theta= -2/sqrt85 and sin theta=9/sqrt85$

Thus $- 2 + 9 i = \sqrt{85} (cos θ + i sin θ)$ $-2 +9i= sqrt85(cos theta +isin theta)$

Like wise -5i +2 would be 2 -5i = $\sqrt{29} (\frac{2}{\sqrt{29}} - i \frac{5}{\sqrt{29}})$ $sqrt29( 2/sqrt 29-i5/sqrt29)$

If $ϕ$ $phi$ is some angle, then let $cos ϕ = \frac{2}{\sqrt{29}} and sin ϕ = - \frac{5}{\sqrt{29}}$ $cos phi= 2/sqrt29 and sin phi= -5/sqrt29$

Thus $2 - 5 i = \sqrt{29} (cos ϕ - sin ϕ)$ $2-5i=sqrt29(cos phi -sin phi)$

$\frac{9 i - 2}{- 5 i + 2} = \sqrt{\frac{85}{29}} \frac{cos θ + i sin θ}{cos ϕ - i sin ϕ} = \sqrt{\frac{85}{29}} \frac{e^{i θ}}{e^{- i ϕ}} = \sqrt{\frac{85}{29}} e^{i (θ + ϕ)}$ $(9i-2)/(-5i+2)= sqrt(85/29) (cos theta +isin theta)/(cosphi-isin phi)=sqrt(85/29) e^(itheta)/e^(-iphi) =sqrt(85/29) e^(i(theta +phi)$

$= \sqrt{\frac{85}{29}} (cos (θ + ϕ) + i sin (θ + ϕ))$ $=sqrt(85/29)( cos(theta+phi)+i sin (theta+phi) )$

$= \sqrt{\frac{85}{29}} [(cos θ cos ϕ - sin θ sin ϕ) + i (sin θ cos ϕ + cos θ sin ϕ)]$ $=sqrt(85/29)[ (cos theta cos phi- sin theta sin phi) +i(sin theta cos phi + cos theta sin phi)]$

$= \sqrt{\frac{85}{29}} [\frac{- 2}{\sqrt{85}} \cdot \frac{2}{\sqrt{29}} - \frac{9}{\sqrt{85}} \cdot \frac{- 5}{\sqrt{29}} + i (\frac{9}{\sqrt{85}} \cdot \frac{2}{\sqrt{29}} + \frac{- 2}{\sqrt{85}} \cdot \frac{- 5}{\sqrt{29}})]$ $=sqrt(85/29)[(-2)/sqrt85 * 2/sqrt29 -9/sqrt85 * (-5)/sqrt29 +i(9/sqrt85 * 2/sqrt29 +(-2)/sqrt85 * (-5)/sqrt29)]$

$\sqrt{\frac{85}{29}} (\frac{41}{\sqrt{85} \sqrt{29}} + i \frac{28}{\sqrt{85} \sqrt{29}})$ $sqrt(85/29) (41/(sqrt85 sqrt29)+i 28/(sqrt 85 sqrt29))$
$= \frac{1}{29} (41 + 28 i)$ $=1/29 (41+28i)$

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How do you divide $\frac{9 i - 2}{- 5 i + 2}$ $( 9i- 2) / (- 5 i+ 2 )$ in trigonometric form?

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How do you divide $\frac{9 i - 2}{- 5 i + 2}$ $( 9i- 2) / (- 5 i+ 2 )$ in trigonometric form?