How do you convert $(- 13, 12)$ $(-13,12)$ to polar form?

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How do you convert $(- 13, 12)$ $(-13,12)$ to polar form?

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Answer 1 · 2017-02-25T14:22:53

$\sqrt{313} [cos (π - arctan \frac{12}{13}) + i sin (π - arctan \frac{12}{13})]$ $sqrt 313 [cos (pi - arctan frac{12}{13}) + i sin (pi - arctan frac{12}{13})]$

Explanation:

$x = r cos t, y = r sin t$ $x = r cos t, y = r sin t$

$- 13 = r cos t, 12 = r sin t$ $-13 = r cos t, 12 = r sin t$

$13^{2} + 12^{2} = r^{2} = 313$ $13^2 + 12^2 = r^2 = 313$

$tan t = - \frac{12}{13}$ $tan t = -12/13$

$t = - arctan \frac{12}{13} + π$ $t = - arctan frac{12}{13} + pi$

$- 13 + 12 i = r (cos t + i sin t)$ $-13 + 12 i = r (cost + i sin t)$

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How do you convert $(- 13, 12)$ $(-13,12)$ to polar form?

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