How do you convert $(3, - 3)$ $(3, -3)$ to polar form?

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Answer 1 · 2016-05-28T02:12:39

$(r, θ) = (3 \sqrt{2}, \frac{7 π}{4})$ $(r, theta)=(3sqrt 2, (7pi)/4)$ or $(3 \sqrt{2}, - \frac{π}{4})$ $(3sqrt 2, -pi/4)$

Use $(x, y) = (3, - 3) = (r cos θ, r sin θ)$ $(x, y) = (3, -3)= (r cos theta, r sin theta)$

$r = \sqrt{x^{2} + y^{2}} = \sqrt{9 + 9} = 3 \sqrt{2}$ $r=sqrt(x^2+y^2)=sqrt(9+9)=3 sqrt 2$

$cos θ = \frac{1}{\sqrt{2}} > 0 and sin θ - \frac{1}{\sqrt{2}} < 0$ $cos theta = 1/sqrt 2>0 and sin theta -1/sqrt2<0$

So, theta is in the 4th quadrant and is $\frac{7 π}{4}$ $(7pi)/4$ and this direction is

same as for $- \frac{π}{4}$ $-pi/4$ .

Now, $(r, θ) = (3 \sqrt{2}, \frac{7 π}{4}))$ $(r, theta)=(3sqrt 2, (7pi)/4))$ or $(3 \sqrt{2}, - \frac{π}{4})$ $(3sqrt 2, -pi/4)$ .

How do you convert (3, -3)(3,−3)(3, -3) to polar form?