# Polar Coordinates, I hate them already

Another useless topic from me, but in light of next year I was advised to look into some topics, and polar coordinates just have me stumped. This is the question I'm not getting.

Sketch the lines (or half-lines) with polar equations:

e) rcos(theta - pi/6) = 2

This is what I did....

Rcos(theta - pi/6) = 2
cos(theta - pi/6) = cos(theta)cos(-pi/6) + sin(theta)sin(-pi.6)

= Rcos(theta)*(root3/2) - 0.5Rsin(theta)

Now: Rcos(theta) = x, Rsin(theta) = y

(root3/2)x - 0.5y = 2 [*2]
root3*x - y = 4
y = root3*x - 4

I get a line, which I think is the answer.

However, take a point (x,y) such as (4, 2.92)

theta = arctan(2.92/4) = 0.631
R = (4^2 + 2.92^2)^0.5 = 4.952

So if we put it back into the original equation:

4.952 * Cos ( 0.631 - pi/6) = 2

However, the left hand side = 4.92, which is totally wrong.

If you take theta as 1.67, this works. Can anybody explain why?

Ad
Maths in the middle of the summer holidays is music to my ears
byb3
Another useless topic from me, but in light of next year I was advised to look into some topics, and polar coordinates just have me stumped. This is the question I'm not getting.

Sketch the lines (or half-lines) with polar equations:

e) rcos(theta - pi/6) = 2

This is what I did....

Rcos(theta - pi/6) = 2
cos(theta - pi/6) = cos(theta)cos(-pi/6) + sin(theta)sin(-pi.6)

= Rcos(theta)*(root3/2) - 0.5Rsin(theta)

Actually,

cos(theta - pi/6)
= cos(theta)cos(-pi/6) - sin(theta)sin(-pi/6) (minus, not plus!)
= cos(theta)cos(pi/6) + sin(theta)sin(pi/6)
= (sqrt(3)/2) cos(theta) + (1/2) sin(theta)

Substituting into r cos(theta - pi/6) = 2 we get

(sqrt(3)/2) x + (1/2) y = 2
ie, y = 4 - sqrt(3) x

Jonny W
cheers mate, i see where I went wrong..

Cos(A-B) = CosACosB + SinASinB

But because I used (-B) on the right, it messed things up!

Thanks for clearing that one up.