Question

How do you find the rectangular coordinates, given the polar coordinates $ ( - 1, - \dfrac{\pi }{6}) $ ?

Answer 1

Hint: In this question, we are given the polar coordinates. A polar coordinate is of the form $ (r,\theta ) $ , so we get the values of r and $ \theta $ , and we have to convert the given polar coordinates into rectangular coordinates, it means that we have to find the value of x and y. We know that $ x = r\cos \theta $ and $ y = r\sin \theta $ , using the value of r and $ \theta $ we will find the value of both x and y and thus get the rectangular coordinates.

Complete step by step solution:
We are given the polar coordinates are $ ( - 1, - \dfrac{\pi }{6}) $ so we get $ r = - 1 $ and $ \theta = - \dfrac{\pi }{6} $
We know that
 $
  x = r\cos \theta \\
   \Rightarrow x = - 1\cos ( - \dfrac{\pi }{6}) \;
  $
We know $ \cos ( - x) = \cos x $
 $
   \Rightarrow x = - \cos \dfrac{\pi }{6} \\
   \Rightarrow x = - \dfrac{{\sqrt 3 }}{2} \;
  $
 $
  y = r\sin \theta \\
   \Rightarrow y = - 1\sin ( - \dfrac{\pi }{6}) \;
  $
We know that $ \sin ( - x) = - \sin x $
 $
   \Rightarrow y = - ( - \sin \dfrac{\pi }{6}) \\
   \Rightarrow y = \dfrac{1}{2} \;
  $
Hence when the polar coordinates are $ ( - 1, - \dfrac{\pi }{6}) $ , the rectangular coordinates are $ ( - \dfrac{{\sqrt 3 }}{2},\dfrac{1}{2}) $ .
So, the correct answer is “$( - \dfrac{{\sqrt 3 }}{2},\dfrac{1}{2}) $ ”.

Note: The rectangular coordinate system is of the form $ (x,y) $ and is the most commonly used coordinate system, where x is the distance of this point from the y-axis and y is the distance of the point from the x-axis. The polar coordinate system is of the form $ (r,\theta ) $ where r is the distance of the point from the origin and $ \theta $ is the counter-clockwise angle between the line joining the point and the origin and the x-axis. Thus, we get a right-angled triangle formed by x, y and r, where r is the hypotenuse, x is the base and y is the height of the triangle, so by Pythagoras theorem, we have - $ {x^2} + {y^2} = {r^2} $ and by trigonometry we have –
 $
  \cos \theta = \dfrac{{base}}{{hypotenuse}} = \dfrac{x}{{\sqrt {{x^2} + {y^2}} }} = \dfrac{x}{r} \\
   \Rightarrow x = r\cos \theta \;
  $
 And similarly $ y = r\sin \theta $