What is the Cartesian form of ( -8 , ( - 15pi)/4 ) (−8,−15π4)?

1 Answer
Psykolord1989 .
Dec 3, 2017

Use the polar conversion formulae x = r cos theta, y = r sin thetax=rcosθ,y=rsinθ. (-4sqrt2, -4sqrt2)(−4√2,−4√2)

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)(r,θ), with rr denoting the distance from the origin, and thetaθ 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)√x2+y2. We can then make a right triangle with rr (the segment) as our hypotenuse, the xx coordinate as the length of one side, and the yy coordinate as the length of the other side.

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

sin theta = y/r, cos theta = x/rsinθ=yr,cosθ=xr

Multiplying both sides by r we get

y = r sin theta, x = r cos thetay=rsinθ,x=rcosθ

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/sqrt2sin(−θ)=−sin(θ),cos(−θ)=cos(θ),sin(θ+2nπ)=sin(θ),cos(θ+2nπ)=cos(θ),sin(π4)=cos(π4)=√22=1√2
Using this, we can perform our conversion:

x = -8cos(-15pi/4) = -8 cos(-7pi/4) = -8cos (7pi/4) = -8sqrt2/2 = -8/sqrt2 = -4sqrt2x=−8cos(−15π4)=−8cos(−7π4)=−8cos(7π4)=−8√22=−8√2=−4√2

y = -8sin(-15pi/4) = -8sin (-7pi/4) = 8 sin(7pi/4) = 8*(-1/sqrt2) = -4sqrt2y=−8sin(−15π4)=−8sin(−7π4)=8sin(7π4)=8⋅(−1√2)=−4√2

The cartesian coordinate form of our point is (x,y) = (-4sqrt2,-4sqrt2)(x,y)=(−4√2,−4√2)

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

r = sqrt(x^2+y^2) = sqrt (32+32) = sqrt(64) = +-8r=√x2+y2=√32+32=√64=±8
theta = tan^-1(y/x) = tan^-1(1) = pi/4 + npiθ=tan−1(yx)=tan−1(1)=π4+nπ, which gives us -7pi/4−7π4 when n=-2n=−2, and -15pi/4−15π4 when n=-4n=−4,

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