polar coordinate question, help?

Question

find out the intersection of p=cos(2theta) and sin(2theta)
the answer on the book is { sqroot(2)/2, (2n+1)Pi/8}

but my answer is different from the book [sqroot(2)/2, Pi/8]
[-sqroot(2)/2, 5Pi/8] [sqroot(2)/2, 9Pi/8] [-sqroot(2)/2, 13Pi/8].....
Note, the only difference between mine and the book is that sqroot(2)/2 alternates between + and - signs.

tell me why i got different answer from the book.

Answer 1

It's just a convention

Some peole measure polar coordinates from 0 to 2π and other people measure them from -π to π.

Some people have r > 0 and others allow -r to be a reflection of r through the origin.

Answer 2

When p = cos2θ and p=sin2θ intersect, then cos2θ = sin2θ.

Square both sides:
cos^2(2θ) = sin^2(2θ).
Rearranging gives cos^2(2θ) - sin^2(2θ) = 0.

Using the trig identity: cos2θ = cos^2θ-sin^2θ
we are left with cos4θ=0.

This is true for 4θ = π/2, 3π/2, 5π/2, etc
so θ = π/8, 3π/8, 5π/8, 7π/8. or (2n+1)π/8 for n = 0, 1, 2, 3...

When 2θ = π/2, sin2θ=sqrt(2)/2

So, the solutions are (p,θ) = ( sqrt(2)/2, (2n+1)π/8 ) for n = 0, 1, 2, 3... just as the book says.

By definition, polar coordinates use only positive radius, so you can discount the negative square root for any values of the radius.

Answer 3

U can not take the square root of a negative no. It has been a very long time but u should have an i on to the square root and then go from there and the i will be in the answer to show that it was of a negative no.

Answer 4

I think that the problem is attributable to the fact you are working with 2O and not plain-old O .