Question

find the rectangular coordinates of the point that has polar coordinates \\(\left(-2, \frac{11\pi}{6}\
ight)\\).
rectangular coordinates: ( , )

Explanation:

Step1: Recall the conversion formulas

The conversion formulas from polar coordinates \((r,\theta)\) to rectangular coordinates \((x,y)\) are \(x = r\cos\theta\) and \(y=r\sin\theta\). Here \(r=- 2\) and \(\theta=\frac{11\pi}{6}\).

Step2: Calculate the \(x\) - coordinate

We know that \(\cos\frac{11\pi}{6}=\cos(2\pi-\frac{\pi}{6})=\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}\).
Using \(x = r\cos\theta\), substitute \(r = - 2\) and \(\cos\theta=\frac{\sqrt{3}}{2}\), we get \(x=-2\times\frac{\sqrt{3}}{2}=-\sqrt{3}\).

Step3: Calculate the \(y\) - coordinate

We know that \(\sin\frac{11\pi}{6}=\sin(2\pi-\frac{\pi}{6})=-\sin\frac{\pi}{6}=-\frac{1}{2}\).
Using \(y = r\sin\theta\), substitute \(r=-2\) and \(\sin\theta =-\frac{1}{2}\), we get \(y=-2\times(-\frac{1}{2}) = 1\).

Answer:

\((-\sqrt{3},1)\)