# Proving Identities

1. Apr 7, 2014

### keishaap

1. The problem statement, all variables and given/known data
Prove that cotx-tanx=2cot 2x
2. Relevant equations

Is 2cot 2x the same as 2(cos2x/sin2x)

3. The attempt at a solution

2. Apr 7, 2014

Yes.

3. Apr 7, 2014

### keishaap

Okay so i get 2(cos^2-sin^2)/(2sinxcosx) when i divide my sins what do i do with the 2 in the denominator?

4. Apr 7, 2014

### LCKurtz

Just write that as two fractions and compare it to the left side.

5. Apr 7, 2014

### keishaap

So its 2(cosx/2sinx)-(sinx/cosx) do the 2 go on both fractions? And i thought you couldnt seperate the fractions unless tgey had the same denominator?

6. Apr 7, 2014

### Staff: Mentor

You have a 2 in the numerator and a 2 in the denominator. You can cancel the 2's. Then split what you have into two fractions.

7. Apr 7, 2014

### 462chevelle

you would want to write (2(cos^2x-sin^2x))/(2sinxcosx) as 2 fractions. the two cancels out.

8. Apr 7, 2014

### keishaap

So would the fractions look like cos^2 x/sinx and sin^2x/cosx ?

9. Apr 7, 2014

### 462chevelle

your whole denominator needs to carry into both fractions

10. Apr 7, 2014

### keishaap

So what would it look like?

11. Apr 7, 2014

### Staff: Mentor

You need to review basic fraction addition and subtraction. One place to start would be khanacademy.org.

12. Apr 7, 2014

### 462chevelle

this.

13. Apr 7, 2014

### keishaap

Lol ever have one of those brain farts where you think things are harder than the seem? Yeah give me a break.

14. Apr 8, 2014

### Staff: Mentor

The questions you asked in the first two quotes above suggest that you're having trouble with the basics, rather than just a momentary lapse. Some review of how to add and multiply fractions would go a long way. If you don't have the mechanics down cold, you absolutely will not be able to complete proofs like the one you posted in this thread.

This isn't meant to be a personal criticism of you - I'm just pointing out an area with some weakness that will keep you from being successful.