# Are the angles of ∏/2 and 3∏/2 coterminal?

1. Dec 15, 2012

### EricPowell

1. The problem statement, all variables and given/known data
Cosx=0. Solve for all angles where 0<x<2∏, and give the general solutions.

2. Relevant equations

3. The attempt at a solution
Angles:
x=∏/2, 3∏/2.

General solutions (what I think they are):
x=∏/2+n2∏
and
x=3∏/x+n2∏
where n is an integer.

General solutions (what the textbook answer key says they are):
x=∏/2+n∏
where n is an integer

Aren't general solutions supposed to give angles that are all coterminal with any integer value of n? If I use n=0, I get ∏/2. And if I use n=1, then I get 3∏/2. I understand that those are both quadrantal angles that lie on the y axis, but they aren't coterminal with one another...are they?

Last edited: Dec 16, 2012
2. Dec 16, 2012

### Mentallic

No, if you use n=2 you have $x=\pi/2+2\pi=5\pi/2$. You would get the solution $x=3\pi/2$ if you used n=1.

And no, $\pi/2$ and $3\pi/2$ are not coterminal angles. A coterminal angle of $\theta$ is any angle that satisfies $x=\theta+2n\pi$ where n is an integer, and $3\pi/2$ does not satisfy this relationship if we let $\theta=\pi/2$.

Keep in mind that two different angles (not coterminal angles) can have the same sine value.

3. Dec 16, 2012

### Staff: Mentor

I don't see how. If n = 2, the textbook answer would be x = $\pi/2 + 2\pi$ = $5\pi/2$, not $3\pi/2$.
By "those angles" are you referring to $\pi/2$ and $3\pi/2$? If so, they are not coterminal.

4. Dec 16, 2012

### EricPowell

Whoops I meant to press the 1 key when I tried to type n=1.

Okay so it seems that I have misunderstood the purpose of general solutions. I thought this whole time that they were meant for producing angles of x that were all coterminal with each other (except for tangent ratios where they only have to be ∏ radians apart) for any integer value of n, but when you reminded me of not co-terminal angles having the same trigonometric ratios, I realised that the textbook's solution of x=∏/2+n∏ will give angles with the same ratios, even though they will not all be coterminal for any integer value of n. My textbook never defined what a general solution was and I made an assumption.

Thank you.

5. Dec 16, 2012

### Mentallic

Yep, that's exactly right. While general solutions can give all coterminal angles, simply because you can tack on $+2n\pi$ at the end, they can do a whole lot more than that. General solutions can be as simple as describing coterminal angles, or as complicated as expressing 10 different angles, and all their coterminal angles into one formula.

The only reason the solution to this problem is so simple ($x=\pi/2+n\pi$) is because you were asked to solve for cos(x)=0 and because of how the cosine wave works, its wavelength is $2\pi$ while it crosses the x-axis or y=0 every $\pi$ units.

If you were instead asked to solve, say, $\cos(x)=1/2$ then the general solution isn't as simple. For $0\leq x<2\pi$ we have $x=\pi/3, 5\pi/3$

And so the coterminal angles would be defined by
$x=\pi/3+2n\pi$
$x=5\pi/3+2n\pi$
as you know, but the general solution which is one formula that provides all solutions to the equation would be

$$x=\frac{\pi}{2}(2n-1)+(-1)^n\frac{\pi}{6}$$

Try it for yourself to see that it works.
Oh and I doubt you need to learn how to construct these more complicated general formulae for a while, so don't stress about it. But if you are curious as to how I did it, take a quadrant and label the principle angles $\pi/3$ and $5\pi/3$ on the circle, now consider this unsimplified formula which is equivalent to the one above:

$$x=n\pi-\frac{\pi}{2}+(-1)^n\left(\frac{\pi}{2}-\frac{\pi}{3}\right)$$

Start at n=1, see how it gives us $x=\pi/3$, then go to n=2 etc.