What is the trigonometric form of $(- 6 - i)$ $(-6-i)$ ?

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What is the trigonometric form of $(- 6 - i)$ $(-6-i)$ ?

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Answer 1 · 2016-03-08T19:12:31

$\sqrt{37} (cos (0.165) + i sin (0.165))$ $sqrt37(cos(0.165) + isin(0.165))$

Explanation:

Using the following formulae:

$∙ r^{2} = x^{2} + y^{2}$ $• r^2 = x^2 + y^2$

$∙ θ = {tan}^{- 1} (\frac{y}{x})$ $• theta = tan^-1(y/x)$

here x = -6 and y = -1

$r^{2} = {(- 6)}^{2} + {(- 1)}^{2} = 37 \Rightarrow r = \sqrt{37}$ $r^2 = (-6)^2+(-1)^2 = 37 rArr r = sqrt37$

and $θ = {tan}^{- 1} (\frac{- 1}{- 6}) = {tan}^{- 1} (\frac{1}{6}) \approx 0.165 radians$ $theta = tan^-1((-1)/(-6)) = tan^-1(1/6) ≈0.165 " radians "$

$\Rightarrow (- 6 - i) = \sqrt{37} (cos (0.165) + i sin (0.165))$ $rArr (-6-i) = sqrt37(cos(0.165) + isin(0.165))$

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What is the trigonometric form of $(- 6 - i)$ $(-6-i)$ ?

1 Answer

Explanation:

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Science

Math

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What is the trigonometric form of (−6−i) (-6-i) ?

1 Answer

Explanation:

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What is the trigonometric form of $(- 6 - i)$ $(-6-i)$ ?