What is the Cartesian form of #( -8 , ( - 15pi)/4 ) #?

1 Answer
Dec 3, 2017

Use the polar conversion formulae #x = r cos theta, y = r sin theta#. #(-4sqrt2, -4sqrt2)#

Explanation:

In order to perform the conversion from polar to cartesian coordinates, it helps us to remind ourselves of what the various coordinates represent.

On the xy-plane, the x coordinate simply denotes the x value (i.e. its position along the x-axis), and the y coordinate denotes the y-value (its position along the y-axis)

The polar coordinates of a given point on the xy-plane, however, are represented by #(r,theta)#, with #r# denoting the distance from the origin, and #theta# the angle formed between the x-axis and the line connecting the given point to the origin.

For a graphical representation, look here:

https://www.mathsisfun.com/polar-cartesian-coordinates.html

Normally we prefer to deal with a positive r-value.
The length of the segment, we know from the distance formula, will be #sqrt(x^2+y^2)#. We can then make a right triangle with #r# (the segment) as our hypotenuse, the #x# coordinate as the length of one side, and the #y# coordinate as the length of the other side.

If the angle formed between the x-axis and r is represented by #theta#, then we know that

#sin theta = y/r, cos theta = x/r#

Multiplying both sides by r we get

#y = r sin theta, x = r cos theta#

Recall several important properties of trigonometric functions to make our conversion process easier:

#sin(-theta) = -sin(theta), cos(-theta) = cos(theta), sin(theta+2npi) = sin(theta), cos(theta+2npi)=cos(theta), sin(pi/4) = cos(pi/4) = (sqrt2)/2 = 1/sqrt2#
Using this, we can perform our conversion:

#x = -8cos(-15pi/4) = -8 cos(-7pi/4) = -8cos (7pi/4) = -8sqrt2/2 = -8/sqrt2 = -4sqrt2#

#y = -8sin(-15pi/4) = -8sin (-7pi/4) = 8 sin(7pi/4) = 8*(-1/sqrt2) = -4sqrt2#

The cartesian coordinate form of our point is #(x,y) = (-4sqrt2,-4sqrt2)#

To confirm, we will convert these back into polar coordinates.

#r = sqrt(x^2+y^2) = sqrt (32+32) = sqrt(64) = +-8#
#theta = tan^-1(y/x) = tan^-1(1) = pi/4 + npi#, which gives us #-7pi/4# when #n=-2#, and #-15pi/4# when #n=-4#,