# goochie1234's questions at Yahoo! Answers regarding polar/rectangular conversions

#### MarkFL

Staff member
Here are the questions:

Trigonometry: polar coordinates help?

1. the rectangular point are given. Find the polar coordinates (R, θ) of this point with θ expressed in radiant. Let R>0 and -2pi< θ<2pi

2. the letters X and Y represent rectangular coordinates, write the given equation using polar coordinates (r,θ)

A. 2r^2 sin (2θ)=1
B.2r^2 cos (2θ)=1
C. 4r^2 sin (2θ)=1
D. 4r^2 cos(2θ)=1

3.the letters X and Y represent rectangular coordinates, write the given equation using polar coordinates (r,θ)

A. r= 5/ cosθ + sin θ
B. r=5
C. r^2=5/ cosθ + sin θ
D. r^2=5

4. the letters r and θ represent polar coordinates. write the given equation using the rectangular coordinates (x,y)

r=8

0=?
I have posted a link there to this topic so the OP can see my work.

#### MarkFL

Staff member
Hello goochie1234,

The key to working these problems are the following relations:

$$\displaystyle (x,y)=(r\cos(\theta),r\sin(\theta))$$

$$\displaystyle x^2+y^2=r^2$$

The first is obtained directly from the definitions of the sine and cosine functions, while the second is a result of the Pythagorean theorem. Consider the following sketch:

We have the point $(x,y)$ to which we draw a line segment from the origin, and lalbe its length as $r$. From the point we drop a vertical line segment to the $x$-axis (at the point $(x,0)$ and its length must be $y$.

Using the definition of the sine function on this right triangle, we find:

$$\displaystyle \sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{y}{r}$$

and so we find:

(1) $$\displaystyle y=r\sin(\theta)$$

Likewise, using the definition of the cosine function on this right triangle, we find:

$$\displaystyle \sin(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{x}{r}$$

and so we find:

(2) $$\displaystyle x=r\cos(\theta)$$

We may also observe that:

$$\displaystyle \frac{y}{x}=\frac{r\sin(\theta)}{r\cos(\theta)}= \tan(\theta)$$

Hence:

(3) $$\displaystyle \theta=\tan^{-1}\left(\frac{y}{x} \right)$$

Note: If $x=0$, then $$\displaystyle \theta=\pm\frac{\pi}{2}$$.

By Pythagoras, we can then easily see:

(4) $$\displaystyle x^2+y^2=r^2$$

So, having these relationships, we can now answer the questions.

1. The rectangular point $(x,y)$ is given. Find the polar coordinates $(r,\theta)$ of this point with $\theta$ expressed in radians. Let $0<r$ and $-2\pi<\theta<2\pi$.

From (3) and (4), we may then state:

$$\displaystyle r=\sqrt{x^2+y^2}$$

$$\displaystyle \theta=\tan^{-1}\left(\frac{y}{x} \right)$$

2.) The letters $x$ and $y$ represent rectangular coordinates. Write the given equations using polar coordinates $(r,\theta)$.

A. $$\displaystyle 2r^2\sin(2\theta)=1$$

I would use the double-angle identity for sine $$\displaystyle \sin(2\alpha)=2\sin(\alpha)\cos(\alpha)$$ to rewrite the equation as:

$$\displaystyle 4r^2\sin(\theta)\cos(\theta)=1$$

We may rewrite this as:

$$\displaystyle 4(r\cos(\theta))(r\sin(\theta))=1$$

Using (1) and (2), we have:

$$\displaystyle 4xy=1$$

B. $$\displaystyle 2r^2\cos(2\theta)=1$$

I would use the double-angle identity for cosine $$\displaystyle \cos(2\alpha)=\cos^2(\alpha)-\sin^2(\alpha)$$ to rewrite the equation as:

$$\displaystyle 2r^2\left(\cos^2(\theta)-\sin^2(\theta) \right)=1$$

Distribute $r^2$:

$$\displaystyle 2\left(r^2\cos^2(\theta)-r^2\sin^2(\theta) \right)=1$$

$$\displaystyle 2\left((r\cos(\theta))^2-(r\sin(\theta))^2 \right)=1$$

Using (1) and (2), we have:

$$\displaystyle 2\left(x^2-y^2 \right)=1$$

C. $$\displaystyle 4r^2\sin(2\theta)=1$$

Referring to part A, we see this will be:

$$\displaystyle 8xy=1$$

D. $$\displaystyle 4r^2\cos(2\theta)=1$$

Referring to part B, we see this will be:

$$\displaystyle 4\left(x^2-y^2 \right)=1$$

3.) The letters $x$ and $y$ represent rectangular coordinates. Write the given equation using polar coordinates $(r,\theta)$.

I am going to make assumptions regarding parts A and C based on experience with giving online help and the lack of bracketing symbols that seems to be prevalent.

A. $$\displaystyle r=\frac{5}{\cos(\theta)+\sin(\theta)}$$

Multiplying through by $$\displaystyle \cos(\theta)+\sin(\theta)$$ we obtain:

$$\displaystyle r\cos(\theta)+r\sin(\theta)=5$$

Using (1) and (2), we have:

$$\displaystyle x+y=5$$

B. $$\displaystyle r=5$$

Since $$\displaystyle 0<r$$ we may square both sides to get:

$$\displaystyle r^2=5^2$$

Using (4), this becomes:

$$\displaystyle x^2+y^2=5^2$$

Since $r=5$ is the locus of all points whose distance is 5 units from the origin, this result should easily follow.

C. $$\displaystyle r^2=\frac{5}{\cos(\theta)+\sin(\theta)}$$

Multiplying through by $$\displaystyle \cos(\theta)+\sin(\theta)$$ we obtain:

$$\displaystyle r\left(r\cos(\theta)+r\sin(\theta) \right)=5$$

Using (1), (2), and (4) there results:

$$\displaystyle \sqrt{x^2+y^2}(x+y)=5$$

D. $$\displaystyle r^2=5$$

Using (4), this becomes:

$$\displaystyle x^2+y^2=5$$

4.) The letters $r$ and $\theta$ represent polar coordinates. Write the given equation using the rectangular coordinates $(x,y)$.

A. $$\displaystyle r=8$$

Square both sides:

$$\displaystyle r^2=8^2$$

Using (4), we have:

$$\displaystyle x^2+y^2=8^2$$

B. $$\displaystyle 0=?$$

I don't know how to interpret this.