Question about converting x = pi/6 to polar form

[ishanz]

Homework Statement

Convert t = pi/6 into polar form.

Homework Equations

x = r*cos(t)
y = r*sin(t)

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?

 

[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.

 

[ishanz]

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

 

[Mentallic]

You're welcome