# Question about converting x = pi/6 to polar form

1. Feb 4, 2010

### ishanz

1. The problem statement, all variables and given/known data
Convert t = pi/6 into polar form.

2. Relevant equations
x = r*cos(t)
y = r*sin(t)

3. The attempt at a solution
t = pi/6
cos(t) = cos(pi/6)
cos(t) = sqrt(3)/2
x/r = sqrt(3)/2
sqrt(3)r = 2x
sqrt(3)sqrt(x^2 + y^2) = 2x
sqrt(3x^2 + 3y^2) = 2x
3x^2 + 3y^2 = 4x^2
0 = x^2 - 3y^2

However, the book says to use tangent instead, as shown below:

t = pi/6
tan t = sqrt(3)/3
y/x = sqrt(3)/3
sqrt(3)x - 3y = 0

This is the correct answer. However, I do not understand why the answer I obtained is incorrect. Is it legal to cosine both sides as I did?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 4, 2010

### Mentallic

Actually the answer you got and the correct answer are nearly the same.

$$\sqrt{3}x-3y=0$$

$$x-\sqrt{3}y=0$$

$$x=\sqrt{3}y$$

$$x^2=3y^2$$

$$x^2-3y^2=0$$

which is what you had. The only problem is that when you squared both sides you have introduced more solutions which are not correct. If we go backwards such as taking the square root of both sides we get:

$$x^2=3y^2$$

$$x=\pm \sqrt{3} y$$

and the solutions $x=-\sqrt{3}y$ are not correct.

It is difficult to avoid squaring in this line you had: $$\sqrt{3(x^2 + y^2)} = 2x$$ which is why using the tangent is the best approach.

3. Feb 4, 2010

### ishanz

Ahh, that makes a lot of sense. Thanks a lot!

4. Feb 4, 2010

### Mentallic

You're welcome