# Precal

posted by on .

How do you find all the polar coordinates of P=(-1, -2pi/3)? I've read that you're supposed to add 2pi to find the coterminals but when I do that I get different answers then in the book. The book answer is (1, 2pi/3+ (2n+ 1) pi) and (-1, -2pi/3+ 2npi). How in the world do you get those answers?

• Precal - ,

you are correct to say that by adding 2π , one rotation , we point ourselves in the same direction,
how about 4π ? , or 6π or n(2π) where n is whole number?
so now the 2nd of their answer should make sense they have simply added multiples of 2π to get
(-1, -2π/3) = (-1, -2π/3 +2nπ)

now look at their 1st answer.
Did you notice that the -1 changed to +1 ?
That would have been a reversal of direction of 180° or π
let n = 0
we get (-1, -2π/3) ---> (+1, 2π/3 + π) or (1, 5π/3)
let's look at this in degrees .....
(-1, -120°) ---> (1, 120 + 180) = (1,300)

At this point we should realize that their 1st answer cannot be correct, since we don't end up at the same endpoint.
(-1, -120°) means: point yourself in the direction of -120
but then go -1 unit, or go 1 unit in the OPPOSITE direction, so the simplest version of
(-1,-120°) = (1,60°) or (1, π/3)

To show it is wrong .....
There has to be an integer value of n such that
2π/3 + (2n+1)π = π/3
2/3 + 2n+1 = 1/3
2n = 4/3
n = 2/3, but n has to be an integer.
so they are wrong.