Problem Verifing Identities in Trig

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Homework Help Overview

The discussion revolves around verifying trigonometric identities, specifically the equation cot²x - 1 / (1 + cot²x) = 2cos²x - 1. Participants are exploring various approaches to prove the identity and clarify their understanding of trigonometric relationships.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss starting from the left side of the equation and applying Pythagorean identities. There are questions about the validity of certain equalities and suggestions to simplify the right side further. Some participants propose using power reduction formulas, while others express confusion about the hints being provided.

Discussion Status

The discussion is active, with participants offering hints and suggestions. Some express frustration over the lack of adherence to these hints, while others acknowledge being on the right track but feeling overwhelmed. There is a mix of interpretations and attempts to clarify the problem without reaching a consensus.

Contextual Notes

Some participants mention not having learned specific formulas, which may limit their approaches. There is also a reference to substituting specific values into the equation to test equality.

Rada3b
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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 someone point me in the right direstion .Thank s in advanced Rada3b
 
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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?
 
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 beginning of the post. Could I simplify the equation on the right anymore that might = cos^2x -sin^2x??
 
Yes you can, but taking Tedjins hint =] Ill make it a tiny bit more obvious - Pythagorean identities?
 
Tedjn said:
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?

Rada3b said:
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 beginning of the post. Could I simplify the equation on the right anymore that might = cos^2x -sin^2x??
"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)?
 
Have you learned about the power reduction formulas yet? It looks easier by using those to me.
 
No, we haven't learned the power reduction formula but it sounds a lot 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.
 
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?
 
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
Gib Z said:
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?

Sorry. I didn't read your posts
 
  • #11
what are the three most basic formulas for proving trig equations?

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  • #12
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.
 

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