# Unit Circle Help

1. Jan 29, 2010

### huntingrdr

1. The problem statement, all variables and given/known data

Find all values of x in the interval [0,2pi] that satisfy the inequality.

sin x > cos x

2. Relevant equations

Unit circle

3. The attempt at a solution

pi/3 > x > 7pi/6

Is that correct?

2. Jan 29, 2010

### Char. Limit

But in your solution, if pi/3 is greater than x, and x is greater than 7pi/6, then pi/3 would be greater than 7pi/6, which isn't true.

First find where sin x = cos x, then use those points and your mind.

3. Jan 29, 2010

### huntingrdr

I know it is everything from pi/6 to 7pi/6, just not sure how to write this as an inequality.

4. Jan 29, 2010

### Staff: Mentor

To represent all the numbers between, say 0 and 3, but not including the endpoints, you can write 0 < x < 3.

5. Jan 29, 2010

### Char. Limit

Also, does sin pi/6 = cos pi/6?

sin of pi/6 is 1/2, but cos of pi/6 is $$\frac{\sqrt{3}}{2}$$.

Thus that can't be one end point, and the same for the other point you have.

6. Jan 29, 2010

### huntingrdr

Alright I think I figured it out. sin x > cos x. pi/3 > x > 7pi/6.

Does that make sense, or is it correct?

7. Jan 29, 2010

### Staff: Mentor

No, and no. pi/3 is not larger than 7pi/6, and neither value is correct for sin x = cos x. You need to find all solutions of this equation in the interval [0, 2pi], and then find the intervals for which sin x > cos x.

8. Jan 29, 2010

### huntingrdr

I am not trying to find where sinx = cosx. I know pi/4, 3pi/4, 5pi/4, and 7pi/4: sinx=cosx though.

Ok I'm not sure how to write it in inequality format but I know starting at pi/3 and going to 7pi/6, the sinx value is great than the cosx value. Right? Now how can I write this in inequality notation?

Maybe it is 7pi/6 > x > pi/3

9. Jan 29, 2010

### Char. Limit

The interval endpoints are wrong. Take the value 2pi/7. The sine of that (.782) is less than the cosine of that (.623), so shouldn't it be in your interval?

10. Jan 29, 2010

### Staff: Mentor

This is your problem, and is what Char. limit and I have been trying to get across to you. In order to find out where sin x > cos x, you need to first find out where sin x = cos x. There are two values of x in [0, 2pi] that satisfy this equation. The interval you specify for which sin x > cos x will have to use these values.
It is not true that sinx = cosx at all of these values. Some of them, yes, but not all of them.
pi/3 and 7pi/6 are totally irrelevant to this problem.

11. Jan 30, 2010

### huntingrdr

OK, so what is the answer? I'm getting no where with this. I don't even see 2pi/7 on the unit circle.

12. Jan 30, 2010

### Staff: Mentor

2pi/7 is the point on the unit circle 2/7 of the way around the upper half of the unit circle, measuring counterclockwise from (1, 0). There's nothing very significant about this number. It was chosen just as an example.

We're not going to give you the answer - that's for you to find - but we'll help you get the answer.

1. Find the values in [0, 2pi] for which sin x = cos x. There are two values, which I will call a and b. You need to find these numbers. The two values you found in the previous step divide the interval [0, 2pi] into three subintervals, [0, a), (a, b), and (b, 2p].
2. Pick a number in the first subinterval. If that number makes sin x > cos x a true statement then every number in that subinterval also makes sin x > cos x a true statement. If the number you choose does not make sin x > cos x a true statement, then no number in that subinterval works, either.
3. Pick a number in the second subinterval. If that number makes sin x > cos x a true statement then every number in that subinterval also makes sin x > cos x a true statement. If the number you choose does not make sin x > cos x a true statement, then no number in that subinterval works, either.
4. Pick a number in the third subinterval. If that number makes sin x > cos x a true statement then every number in that subinterval also makes sin x > cos x a true statement. If the number you choose does not make sin x > cos x a true statement, then no number in that subinterval works, either.
The solution is the union of the subintervals for which sin x > cos x.

13. Jan 30, 2010

### huntingrdr

Ok, I split it up like you said. I have [0, pi/4), (pi/4, 5pi/4), and (5pi/4,2pi].

1. On the interval [0,pi/4) cos x is > sin x so this case does not work.
2. On the interval (pi/4, 5pi/4) sinx > cosx, so this is a true statement.
3. On the interval (5pi/4, 2pi] cosx > sin x so this case does not work.

So the interval is 5pi/4 < x < pi/4 ?

14. Jan 30, 2010

### Staff: Mentor

Almost. The interval is pi/4 < x < 5pi/4, or in interval notation (pi/4, 5pi/4). This is one way of saying that pi/4 < x AND x < 5pi/4.

What's wrong with the way you wrote it is that 5pi/4 > pi/4, which is not true. You're also saying that x > 5pi/4, and the x < pi/4. You've already shown that those intervals don't work.

15. Jan 30, 2010

### huntingrdr

Thanks.

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