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

What is the cartesian form of (3,5π2)\left( 3,\dfrac{-5\pi }{2} \right) ?

Explanation

Solution

We need to find the cartesian form of (3,5π2)\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=rcosθx=r\cos \theta and y=rsinθy=r\sin \theta where r=3 and θ=5π2\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=rcosθx=r\cos \theta and y=rsinθy=r\sin \theta
The polar coordinate system, in mathematics, is a two-dimensional coordinate system. It is usually represented as (r,θ)\left( r,\theta \right)
r is the distance between the point and the fixed origin
θ\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 (x,y)\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,
(r,θ)=(3,5π2)\Rightarrow \left( r,\theta \right)=\left( 3,\dfrac{-5\pi }{2} \right)
From the above, we have
r = 3;
θ\theta = 5π2\dfrac{-5\pi }{2}
The above polar coordinates can be converted into cartesian coordinates as follows,
x=rcosθ\Rightarrow x=r\cos \theta
Substituting the value of r and θ\theta , we get,
x=3cos(5π2)\Rightarrow x=3\cos \left( \dfrac{-5\pi }{2} \right)
From trigonometry, we know that cos(θ)=cosθ\cos \left( -\theta \right)=\cos \theta
Writing the same, we get,
x=3cos(5π2)\Rightarrow x=3\cos \left( \dfrac{5\pi }{2} \right)
x=3cos(2π+π2)\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 1st{{1}^{st}} quadrant.
Simplifying the above equation, we get,
x=3cos(π2)\Rightarrow x=3\cos \left( \dfrac{\pi }{2} \right)
From trigonometry, we know that cos(π2)=0\cos \left( \dfrac{\pi }{2} \right)=0
Substituting the same, we get,
x=3×0\Rightarrow x=3\times 0
x=0\therefore x=0

y=rsinθ\Rightarrow y=r\sin \theta
Substituting the value of r and θ\theta , we get,
y=3sin(5π2)\Rightarrow y=3\sin \left( \dfrac{-5\pi }{2} \right)
From trigonometry, we know that sin(θ)=sinθ\sin \left( -\theta \right)=-\sin \theta
Writing the same, we get,
y=3sin(5π2)\Rightarrow y=-3\sin \left( \dfrac{5\pi }{2} \right)
y=3sin(2π+π2)\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 1st{{1}^{st}} quadrant.
Simplifying the above equation, we get,
y=3sin(π2)\Rightarrow y=-3\sin \left( \dfrac{\pi }{2} \right)
From trigonometry, we know that sin(π2)=1\sin \left( \dfrac{\pi }{2} \right)=1
Substituting the same, we get,
y=3×1\Rightarrow y=-3\times 1
y=3\therefore y=-3
\therefore The cartesian form of (3,5π2)\left( 3,\dfrac{-5\pi }{2} \right) is (0,3)\left( 0,-3 \right)

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