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

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SUMMARY

The identity cos(2A)/(1 + sin(2A)) = (cot(A) - 1)/(cot(A) + 1 has been successfully proven by multiple participants in the discussion. The right side simplifies to (cos(A) - sin(A))/(cos(A) + sin(A)), which aligns with the left side when manipulated correctly. Key techniques included the use of double angle formulas and algebraic manipulation, particularly the identity sin^2(A)/(sin^2(A) + sin^2(A)) = 1. Participants emphasized the importance of careful fraction handling and the application of algebraic identities.

PREREQUISITES
  • Understanding of trigonometric identities, specifically cotangent and sine functions.
  • Familiarity with double angle formulas, such as sin(2A) = 2sin(A)cos(A) and cos(2A) = cos^2(A) - sin^2(A).
  • Proficiency in algebraic manipulation, including factoring and simplifying fractions.
  • Knowledge of basic mathematical identities, such as (a + b)^2 = a^2 + 2ab + b^2 and a^2 - b^2 = (a - b)(a + b).
NEXT STEPS
  • Study the derivation and applications of double angle formulas in trigonometry.
  • Practice simplifying complex trigonometric identities using algebraic techniques.
  • Explore the properties and transformations of cotangent functions in various contexts.
  • Review advanced algebraic identities and their proofs to enhance mathematical reasoning skills.
USEFUL FOR

Students, educators, and anyone involved in mathematics, particularly those focusing on trigonometry and algebraic manipulation of identities.

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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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