How do you convert $(- \sqrt{2}, 3 \frac{π}{4})$ $(-sqrt2, 3pi/4)$ to rectangular form?

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Answer 1 · 2016-03-10T12:41:22

$(- \sqrt{2}, \frac{3 π}{4})$ $(-sqrt2,(3pi)/4)$ in rectangular coordinates is $(1, - 1)$ $(1,-1)$

$(r, θ)$ $(r,theta)$ in polar coordinates is $(r cos θ, r sin θ)$ $(rcostheta,rsintheta)$ in rectangular coordinates.

Hence, $(- \sqrt{2}, \frac{3 π}{4})$ $(-sqrt2,(3pi)/4)$ in rectangular coordinates is

$(- \sqrt{2} \times cos (\frac{3 π}{4}), - \sqrt{2} \times sin (\frac{3 π}{4}))$ $(-sqrt2xxcos((3pi)/4),-sqrt2xxsin((3pi)/4))$ or

$((- \sqrt{2}) \times (- \frac{\sqrt{2}}{2}), (- \sqrt{2}) \times \frac{\sqrt{2}}{2})$ $((-sqrt2)xx(-sqrt2/2),(-sqrt2)xxsqrt2/2)$ or

$((\frac{2}{2}), - (\frac{2}{2}))$ $((2/2),-(2/2))$ or

$(1, - 1)$ $(1,-1)$

How do you convert (-sqrt2, 3pi/4) (−√2,3π4)(-sqrt2, 3pi/4) to rectangular form?