How do you divide $\frac{2 i - 1}{- i - 1}$ $( 2i-1) / ( -i -1 )$ in trigonometric form?

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Answer 1 · 2018-07-02T11:53:26

$\sqrt{\frac{5}{2}} ({cos 108.435}^{\circ} - {sin 108.435}^{\circ})$ $\sqrt{5/2}(\cos108.435^\circ-\sin 108.435^\circ)$

Given that

$\frac{2 i - 1}{- i - 1}$ $\frac{2i-1}{-i-1}$

$= \frac{1 - 2 i}{1 + i}$ $=\frac{1-2i}{1+i}$

$= \frac{(1 - 2 i) (1 - i)}{(1 + i) (1 - i)}$ $=\frac{(1-2i)(1-i)}{(1+i)(1-i)}$

$= \frac{1 - 3 i + 2 i^{2}}{1 - i^{2}}$ $=\frac{1-3i+2i^2}{1-i^2}$

$= \frac{1 - 3 i + 2 (- 1)}{1 + 1}$ $=\frac{1-3i+2(-1)}{1+1}$

$= \frac{- 1 - 3 i}{2}$ $=\frac{-1-3i}{2}$

$= - \frac{1}{2} - \frac{3}{2} i$ $=-1/2-3/2i$

Amplitude $= \sqrt{{(- \frac{1}{2})}^{2} + {(- \frac{3}{2})}^{2}} = \sqrt{\frac{5}{2}}$ $=\sqrt{(-1/2)^2+(-3/2)^2}=\sqrt{5/2}$

Argument, $θ = - (π - {tan}^{- 1} ∣ ∣ ∣ ∣ \frac{- \frac{3}{2}}{- \frac{1}{2}} ∣ ∣ ∣ ∣)$ $\theta=-(\pi-\tan^{-1}|\frac{-3/2}{-1/2}|)$

$θ = - {108.435}^{\circ}$ $\theta=-108.435^\circ$

$\therefore \frac{2i-1}{-i-1}$

$=\sqrt{5/2}(\cos(-108.435^\circ)+i\sin(-108.435^\circ))$

$=\sqrt{5/2}(\cos108.435^\circ-\sin 108.435^\circ)$

How do you divide ( 2i-1) / ( -i -1 )2i−1−i−1( 2i-1) / ( -i -1 ) in trigonometric form?