Question

What is the cartesian form of $\left( 3,\dfrac{-5\pi }{2} \right)$ ?

Answer 1

Hint: We need to find the cartesian form of $\left( 3,\dfrac{-5\pi }{2} \right)$ . We need to find the cartesian coordinates for the given polar coordinates. We find the cartesian coordinates using the formulae $x=r\cos \theta$ and $y=r\sin \theta$ where r=3 and $\theta =\dfrac{-5\pi }{2}$

Complete step by step solution:
We are given polar coordinates and are asked to find the cartesian coordinates for the same. We will be solving the given question using the formulae $x=r\cos \theta$ and $y=r\sin \theta$
The polar coordinate system, in mathematics, is a two-dimensional coordinate system. It is usually represented as $\left( r,\theta \right)$
r is the distance between the point and the fixed origin
$\theta$ is the angle between the point and the fixed direction
We know that the cartesian coordinates of a point are a pair of numbers that have a specific signed distance from the coordinate axis. We can also refer to the cartesian coordinates as rectangular coordinates.
The cartesian coordinates plane consists of two-axis perpendicular to each other.
We know that the horizontal axis is referred to as the x-axis and the vertical axis is referred to as the y-axis.
We can say that the origin is the intersection of the x-axis and y-axis.
We can represent the coordinates of the cartesian plane as $\left( x,y \right)$.
Here, we have the first number x which denotes the distance of the point along the x-axis and the second number y which denotes the distance of the point along the y-axis.
According to our question,
$\Rightarrow \left( r,\theta \right)=\left( 3,\dfrac{-5\pi }{2} \right)$
From the above, we have
r = 3;
$\theta$ = $\dfrac{-5\pi }{2}$
The above polar coordinates can be converted into cartesian coordinates as follows,
$\Rightarrow x=r\cos \theta$
Substituting the value of r and $\theta$ , we get,
$\Rightarrow x=3\cos \left( \dfrac{-5\pi }{2} \right)$
From trigonometry, we know that $\cos \left( -\theta \right)=\cos \theta$
Writing the same, we get,
$\Rightarrow x=3\cos \left( \dfrac{5\pi }{2} \right)$
$\Rightarrow x=3\cos \left( 2\pi +\dfrac{\pi }{2} \right)$
The value of the cosine function lies in the first quadrant. The value of the cosine function is positive in the ${{1}^{st}}$ quadrant.
Simplifying the above equation, we get,
$\Rightarrow x=3\cos \left( \dfrac{\pi }{2} \right)$
From trigonometry, we know that $\cos \left( \dfrac{\pi }{2} \right)=0$
Substituting the same, we get,
$\Rightarrow x=3\times 0$
$\therefore x=0$

$\Rightarrow y=r\sin \theta$
Substituting the value of r and $\theta$ , we get,
$\Rightarrow y=3\sin \left( \dfrac{-5\pi }{2} \right)$
From trigonometry, we know that $\sin \left( -\theta \right)=-\sin \theta$
Writing the same, we get,
$\Rightarrow y=-3\sin \left( \dfrac{5\pi }{2} \right)$
$\Rightarrow y=-3\sin \left( 2\pi +\dfrac{\pi }{2} \right)$
The value of the sine function lies in the first quadrant. The value of the sine function is positive in the ${{1}^{st}}$ quadrant.
Simplifying the above equation, we get,
$\Rightarrow y=-3\sin \left( \dfrac{\pi }{2} \right)$
From trigonometry, we know that $\sin \left( \dfrac{\pi }{2} \right)=1$
Substituting the same, we get,
$\Rightarrow y=-3\times 1$
$\therefore y=-3$
$\therefore$ The cartesian form of $\left( 3,\dfrac{-5\pi }{2} \right)$ is $\left( 0,-3 \right)$

Note: The value of the cosine function is positive in the ${{1}^{st}}$ and the ${{4}^{th}}$ quadrants and is negative in the ${{2}^{nd}}$ and the ${{3}^{rd}}$ quadrant. The value of the sine function is positive in the ${{1}^{st}}$ and the ${{2}^{nd}}$ quadrants and is negative in the ${{3}^{rd}}$ and the ${{4}^{th}}$ quadrants.