# Trig pi question

1. May 27, 2010

### DrummingAtom

1. The problem statement, all variables and given/known data
If pi ≤ Θ ≤ 2pi and cos Θ = cos 1, what is the value of Θ? Round to nearest hundredth.

2. Relevant equations

3. The attempt at a solution
cos 1 = ~.54 then I didn't really understand how to interpret the "If pi ≤ Θ ≤ 2pi" Thanks for your help.

2. May 27, 2010

### Mentallic

Well since $cos(\theta)=cos(1)$ then this would instinctively mean $\theta=1$ but the restriction is $\pi \leq \theta \leq 2\pi$ or $3.14 \leq \theta \leq 6.28$ (approx). Obviously $1<\pi$ so we can't use the instinctive solution.

What other values of $\theta$ make the same value cos(1)? There are infinite values of $\theta$ that do this. Take a look at the cosine graph and find where $\theta=1$ (or x=1). Now the y-value at that point is cos(1). Where else does the same y value occur between $\pi < \theta< 2\pi$?

3. Jun 12, 2010

### Axiom17

Apologies for the rather crude diagram (follow the link below) but hopefully this will help along with the advice in the previous post.

http://yfrog.com/6fpf1kj

4. Jun 12, 2010

### cronxeh

180 deg <= Θ <= 360 deg

cos Θ = cos 1

cos 359 = cos 1

Particularly,
Θ= 2*pi*n - 1
Θ=2*180*n - 1 = 360*n - 1
Θ= 359, 719, 1079, 1439, etc degrees
180 < 359 < 360 for this case

Last edited: Jun 12, 2010
5. Jun 12, 2010

### Axiom17

Think about this.. the cos graph in the region $0 \leq \pi \leq 2\pi$ is symmetrical about $\pi$.

At $x=1$ you have $y=cos(1)$, and you need to find the other point, i.e. at a different x value, where you get the same y value.

Perhaps if you knew the difference from the lower limit to the first x value, you could say this was the same as from the upper limit to the value you require? if that makes sense, look at the graph again.

Also, you should be able to see that $cos(1)\neq cos(359)$

6. Jun 12, 2010

### cronxeh

heh. Ok enlighten us, for which value of 180<x<360 is cos(x)=cos(1) ?

And dont tell me its cos 5.283185 = cos 1. Nobody said we doing this in radians

Last edited: Jun 12, 2010
7. Jun 12, 2010

### Axiom17

.. hah yeah, what's happened there, is that I've done some bad maths! How strange!

Hopefully my method made some sense though. I've been working in radians not degrees as that's what was stated in the question. So what I did was:

$$cos(1)=0.54=cos(2\pi -1)=cos(5.28)$$

So $\theta=5.28 radians=302.7 degrees$

.. Perhaps I need a refresher in Trig! ..never mind..