# Homework Help: Problem Verifing Identities in Trig

1. Mar 6, 2008

Hello this one has stumped me for about a day and I have tried several different approaches. maybe someone could point me in the right direction.
cot^2x-1/1+cot^2x=2Cos^2x-1

I startedon the Left side tring to prove the equation.

cot^2x-1/csc^2x using Pyth. Theory

(cot^2x/csc^2x)-(1/csc^2x) reciprocal Identites

((cos^2x/sin^2x)/csc^2x)-(sin^2x) reciprocal Identites

(cos^2x/sin^2x)*(sin^2x)-(sin^2x) //sin^2x will cancel outleaving

cos^2x - sin^2x this does not = 2Cos^2x -1

Could somone point me in the right direstion .Thank s in advanced Rada3b

2. Mar 6, 2008

### Tedjn

Are you sure that

$$\cos^2x - \sin^2x = 2\cos^2x - 1$$

is not a true statement? What is 1 equal to in terms of trig identities?

3. Mar 7, 2008

You are correct it is not. I was trying to prove that the equation on the left is = to the equation on the right in the begining of the post. Could I simplify the equation on the right anymore that might = cos^2x -sin^2x??

4. Mar 7, 2008

### Gib Z

Yes you can, but taking Tedjins hint =] Ill make it a tiny bit more obvious - Pythagorean identities?

5. Mar 7, 2008

### HallsofIvy

"You are correct it is not"? Tedjn was not telling you it is not true! He was suggesting that you check your assumption that it is not true again! Specifically, what do you get if you replace $sin^2(x)$ on the left side by $1- cos^2(x)$?

6. Mar 7, 2008

### Hydrargyrum

Have you learned about the power reduction formulas yet? It looks easier by using those to me.

7. Mar 8, 2008

No, we haven't learned the power reduction formula but it sounds alot easier Hydrargyrum. I will look into it and continue trying. I did sub Pi/4 in for x on both side ofthe equation and yes they are =. So back to proving again, Thanks EV1 for the help.

8. Mar 8, 2008

### Gib Z

Would you PLEASE listen to any of our hints? The power reduction formulas come from what we are trying to tell you. I don't see why our hint, which has become more so an obvious instruction, is so hard to execute?

9. Mar 8, 2008

Thank you all. I was on the right track.I must be just overstressed atm not to see something so obvious. I got it now.

10. Mar 9, 2008

### Hydrargyrum

11. Mar 9, 2008

### Charlie_russo

what are the three most basic formulas for proving trig equations?

blank+blank=1
1-blank=blank
1-blank=blank

12. Mar 24, 2008

### Atheleos

L= cot^2x-1/csc^2x =cot^2x/csc^2x -1/csc^2x =(cos^2x/sin^2x)/(1/sin^2x)-sin^2x=cos^2x-sin^2x=cos^2x-(1-cos^2x)=cos^2x-1+ cos^2x=2cos^2x-1=R

Everybody gave you hints. Its been a while now. This should be your answer. Please correct if I am wrong.