Understanding the Simplification of ((1 - cos A)/ (1+cos A) ) ^(1/2)

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SUMMARY

The discussion focuses on proving the identity \(\sqrt{\frac{1 - \cos A}{1 + \cos A}} = \csc A - \cot A\). Participants clarify that the expression can be simplified by recognizing that \(1 - \cos^2 A = \sin^2 A\), leading to the transformation of the original expression into \(\frac{1 - \cos A}{\sin A}\). This ultimately results in the definitions of cosecant and cotangent, confirming the identity. Key steps include manipulating the numerator and denominator appropriately and applying fundamental trigonometric identities.

PREREQUISITES
  • Understanding of basic trigonometric identities, including \(\sin^2 A + \cos^2 A = 1\)
  • Familiarity with the definitions of cosecant (\(\csc A\)) and cotangent (\(\cot A\))
  • Ability to manipulate algebraic fractions and square roots
  • Knowledge of how to simplify expressions involving trigonometric functions
NEXT STEPS
  • Study the derivation and applications of trigonometric identities in proofs
  • Learn about the properties of square roots in algebraic expressions
  • Explore advanced trigonometric identities, such as the Pythagorean identities
  • Practice simplifying complex trigonometric expressions using various identities
USEFUL FOR

Students studying trigonometry, educators teaching trigonometric identities, and anyone looking to enhance their understanding of algebraic manipulation in trigonometric contexts.

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Homework Statement


Prove:( (1 - cos A)/ (1+cos A) ) ^(1/2) = cosec A - cot A

Then they have...

((1 - cos A)/ (1+cos A) ) ^(1/2) = ( (1 - cos A)^2 )/ (1-cos^2 A) ) ^(1/2)

= (1 - cos A) / (1-cos^2 A) ^(1/2)/

I have never done anything like this... I have just been studying trig for the past two days...

Why did they square the entire numerator but only square the cos in the denominator?

... Can some help me understand this the easiest way possible? thank you.


Homework Equations





The Attempt at a Solution

 
Last edited by a moderator:
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Nope lol sorry... However I did figure it out on my own by just looking at it! So I was proud about that lol...

One important question that I would like answered...
Lets say... I have... (1 - cos A)/ (1+cos A) ) ^(1/2)
Should I leave it as so... or try and simplify it down to cosec A - cot A?
Because to be honest I don't think I could have looked at the square root portion and though " Ohh... maybe this can be simplified..."
 
Last edited:
Miike012 said:

Homework Statement


Prove:( (1 - cos A)/ (1+cos A) ) ^(1/2) = cosec A - cot A

Then they have...

((1 - cos A)/ (1+cos A) ) ^(1/2) = ( (1 - cos A)^2 )/ (1-cos^2 A) ) ^(1/2)

= (1 - cos A) / (1-cos^2 A) ^(1/2)
Once you are here, use the fact that 1- cos^2(A)= sin^2(A)
(From sin^2(A)+ cos^2(A)= 1. Do you know that identity?)

\left(\frac{1- cos(A)}{1- cos^2(A)}\right)^{1/2}= \frac{1- cos(A)}{sin(A)}= \frac{1}{sin(A)}- \frac{cos(A)}{sin(A)}
Do you know the definition of "cosec(A)" and "cot(A)"?

I have never done anything like this... I have just been studying trig for the past two days...

Why did they square the entire numerator but only square the cos in the denominator?
They didn't. What they did is multiply both numerator and denominator by 1- cos(A). Since there was already 1- cos(A), that becomes (1- cos(A))^2. The denominator was 1+ cos(A) so it becomes (1- cos(A))(1+ cos(A))= 1+ 1(cos(A))- cos(A)(1)- cos(A)cos(A)= 1- cos^2(A) because the "1(cos(A))"and "[mat]-cos(A)(1)[/itex] cancel.

... Can some help me understand this the easiest way possible? thank you.


Homework Equations





The Attempt at a Solution

 
Last edited by a moderator:
Ive been reading in my book.. and it says,

sin^2 A + cos^2 A = 1
Sec^2A = 1 + tan^2A
.
.
. Should I also rearange sin^2 A + cos^2 A = 1 ; cos^2 A = 1 -sin^2 A = sin^2 A =1 -cos^2..
and the same with the other one?
Should I not only memorize the two but also memorize their rearangements?
 

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