Question

Plot the point whose polar coordinates are given. Then find the Cartesian coordinates of the point.

\(\begin{aligned}{l}(a)( - \sqrt 2 ,5\pi /4)\\(b)(1,5\pi /2)\\(c)(2, - 7\pi /6)\end{aligned}\)

Short answer

(a)\((x,y) = (1,1)\)

(b)\((x,y) = (0,1)\)

(c)\((x,y) = ( - \sqrt 3 ,1)\)

Step-by-step solution

Convert polar coordinate to rectangular coordinate

\(\begin{aligned}{l}x &= r\cos \theta = - \sqrt 2 \cos \left[ {\frac{{5\pi }}{4}} \right] = \sqrt 2 \cos \left[ {\frac{{5\pi }}{4} - \pi } \right] = \sqrt 2 \cos \left[ {\frac{\pi }{4}} \right] = \sqrt 2 \times \frac{1}{{\sqrt 2 }} = 1\\y = r\sin \theta &= - \sqrt 2 \sin \left[ {\frac{{5\pi }}{4}} \right] = \sqrt 2 \sin \left[ {\frac{{5\pi }}{4} - \pi } \right] = \sqrt 2 \sin \left[ {\frac{\pi }{4}} \right] = \sqrt 2 \times \frac{1}{{\sqrt 2 }} = 1\end{aligned}\)

Remember that:

\(\begin{aligned}{l}\sin (\pi - \theta ) &= \sin \theta \\\cos (\pi - \theta ) &= - \cos \theta \end{aligned}\)

The point\((1,1)\) is plotted on the graph below:

Convert polar coordinate to rectangular coordinate

\(\begin{aligned}{l}x &= r\cos \theta = 2\cos \left[ { - \frac{{7\pi }}{6}} \right] = - 2\cos \left[ {\pi - \frac{{7\pi }}{6}} \right] = - 2\cos \left[ { - \frac{\pi }{6}} \right] = - 2\cos \left[ {\frac{\pi }{6}} \right] = - 2 \times \frac{{\sqrt 3 }}{2} = - \sqrt 3 \\y &= r\sin \theta = 2\sin \left[ { - \frac{{7\pi }}{6}} \right] = - 2\sin \left[ {\pi - \frac{{7\pi }}{6}} \right] = - 2\sin \left[ { - \frac{\pi }{6}} \right] = 2\sin \left[ {\frac{\pi }{6}} \right] = 2 \times \frac{1}{2} = 1\end{aligned}\)

Remember that:

\(\begin{aligned}{l}\sin (\pi + \theta ) &= - \sin \theta \\\cos (\pi + \theta ) &= - \cos \theta \end{aligned}\)

And that:

\(\begin{aligned}{l}\sin ( - \theta ) &= - \sin \theta \\\cos ( - \theta ) &= \cos \theta \end{aligned}\)

The point\(( - \sqrt 3 ,1)\) is plotted on the graph below:

Convert polar coordinate to rectangular coordinate

\(\begin{aligned}{l}x &= r\cos \theta = 2\cos \left[ { - \frac{{7\pi }}{6}} \right] = - 2\cos \left[ {\pi - \frac{{7\pi }}{6}} \right] = - 2\cos \left[ { - \frac{\pi }{6}} \right] = - 2\cos \left[ {\frac{\pi }{6}} \right] = - 2 \times \frac{{\sqrt 3 }}{2} = - \sqrt 3 \\y &= r\sin \theta = 2\sin \left[ { - \frac{{7\pi }}{6}} \right] = - 2\sin \left[ {\pi - \frac{{7\pi }}{6}} \right] = - 2\sin \left[ { - \frac{\pi }{6}} \right] = 2\sin \left[ {\frac{\pi }{6}} \right] = 2 \times \frac{1}{2} = 1\end{aligned}\)Remember that:

\(\begin{aligned}{l}\sin (\pi + \theta ) &= - \sin \theta \\\cos (\pi + \theta ) &= - \cos \theta \end{aligned}\)

And that:

\(\begin{aligned}{l}\sin ( - \theta ) &= - \sin \theta \\\cos ( - \theta ) &= \cos \theta \end{aligned}\)

The point\(( - \sqrt 3 ,1)\)is plotted on the graph below: