Proving Identity: cos2A/1 + sin2A = cotA - 1 / cotA + 1

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Homework Help Overview

The discussion revolves around proving the trigonometric identity: cos2A/(1 + sin2A) = (cotA - 1)/(cotA + 1). Participants are exploring various approaches to establish the validity of this identity within the context of trigonometric functions.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • One participant attempts to prove the right side of the identity but struggles with the left side, expressing frustration over multiple failed attempts. Others suggest using double angle formulas and question the derivation of certain expressions, such as sin^2A(cot^2 - 1). There are inquiries about the manipulation of fractions and the use of common factors in the proof process.

Discussion Status

Some participants have reported progress in their attempts, with one claiming to have solved the identity, while others are still exploring different methods and questioning the steps taken. There is a mix of interpretations and approaches being discussed, indicating an ongoing exploration of the problem.

Contextual Notes

Participants are navigating through various algebraic manipulations and trigonometric identities, with some expressing confusion over specific steps and the use of mathematical properties. The discussion reflects a learning environment where assumptions and methods are being critically examined.

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Homework Statement


Prove the following identity:

cos2A/1 + sin2A = cotA - 1 / cotA + 1



Homework Equations





The Attempt at a Solution



I proved the right side, which eventually lead up to cosA - sinA / cosA + sinA

I have NO idea how to do the left side. I have wasted roughly 10-15 sheets of paper now trying to prove it, but it all leads up to nothing. I tried MANY different solutions...
 
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did u use the double angle formulas of sin2u=2sinucosu and cos2u=1-2sin^2u?
 
http://img141.imageshack.us/img141/8388/trigom8.jpg
 
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Vitaly, thanks for trying BUT I have a question about where sin^2a(cot^2 - 1) came from.. could you add me to MSN please? supaflygt@hotmail.com
 
http://img139.imageshack.us/img139/9676/trigkw2.jpg
no probs :wink:
Look carefully, it's sine squared, not sin(2A). My 2A are big
 
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Yeah, I know. I'm curious how you got sin^2a/sin^2a + sin^2a in the equation (2nd step) :P
 
Please use brackets where necessary. (sinA)^2/(sinA)^2=1 and as you must know if you multiply something by 1, it doesn't change anything. It's a very common trick in math, e.g. integrating logs, taking inverse of a matrix, etc..
So, I muultipled (cosA)^2 by 1. And (sinA)^2 is therefore a common factor.
It's like (5+7)=(5*7/7+7)=7(5/7+1)
 
Well, I tried it doing my own thing, and here is the result:
EDIT: it's not -1 or 0, but it must be something.
2e1d1lg.jpg
 
Solved it now, thanks for your help.
 
  • #10
Draggu said:
Solved it now, thanks for your help.

my friend they are not equal to negative one. however you can prove both sides are
(cos A-sin A)/(cos A+sin A)
 
  • #11
My God, seeing those divisions just scares me.
Revisit fractions please and formulas of (a+b)2, a2-b2

And consider this: (52-32)/(5+3)2 is NOT equal -1.
(52-32)/(5+3)2=(5-3)(5+3)/((5+3)(5+3)). Divide by a common factor of 5+3, leaving you with (5-3)/(5+3)

Learn (a+b)2=(a+b)(a+b)=a2+2ab+b2, a2-b2=(a-b)(a+b) NOW
 
  • #12
-Vitaly- said:
My God, seeing those divisions just scares me.
Revisit fractions please and formulas of (a+b)2, a2-b2

And consider this: (52-32)/(5+3)2 is NOT equal -1.
(52-32)/(5+3)2=(5-3)(5+3)/((5+3)(5+3)). Divide by a common factor of 5+3, leaving you with (5-3)/(5+3)

Learn (a+b)2=(a+b)(a+b)=a2+2ab+b2, a2-b2=(a-b)(a+b) NOW

I know they aren't equal to -1. I proved both sides were (cosa-sina)/(cosa+sina), though, which was why I said it was solved
 
  • #13
Draggu said:
I know they aren't equal to -1. I proved both sides were (cosa-sina)/(cosa+sina), though, which was why I said it was solved

He says so because of your amazing cancellation.
 

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