Pretty difficult trig proof (identity)

In summary, the problem involves simplifying the expression \frac{sin\theta}{1-cos\theta} - \frac{cot\theta}{1+cos\theta} = \frac{1-cos^{3}\theta}{sin^{3}\theta} using trigonometric identities. The original poster attempted to simplify it to the expression \frac{sin\theta+(cos^{2}\theta)(sin\theta)}{sin^{2}\theta}, but this is incorrect as it does not account for the cot function. The correct solution involves factoring out a common factor of sin(theta) in the numerator and simplifying from there.
  • #1
iRaid
559
8

Homework Statement


[itex]\frac{sin\theta}{1-cos\theta} - \frac{cot\theta}{1+cos\theta} = \frac{1-cos^{3}\theta}{sin^{3}\theta}[/itex]


Homework Equations


Trig identities..


The Attempt at a Solution


Basically I got to:
[itex]\frac{sin\theta+(cos^{2}\theta)(sin\theta)}{sin^{2}\theta}[/itex]

Homework Statement



Is that right up to there, I think not because I can't get passed this lol.


Homework Equations





The Attempt at a Solution

 
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  • #2
That doesn't look right.

Show how you got your result so we can help you.
 
  • #3
iRaid said:

Homework Statement


[itex]\frac{sin\theta}{1-cos\theta} - \frac{cot\theta}{1+cos\theta} = \frac{1-cos^{3}\theta}{sin^{3}\theta}[/itex]


Homework Equations


Trig identities..


The Attempt at a Solution


Basically I got to:
[itex]\frac{sin\theta+(cos^{2}\theta)(sin\theta)}{sin^{2}\theta}[/itex]

Homework Statement



Is that right up to there, I think not because I can't get passed this lol.


Homework Equations





The Attempt at a Solution


You did notice the cot function did you - or did you misread it as cos?
 
  • #4
I don't think you were supposed to solve the question for the Original Poster!
 
Last edited by a moderator:
  • #5
>_> Oh. Well, I hope my explanation blurb thing helps so that I'm not just blatantly giving the solution without providing any real understanding.
 
  • #6
Dr. Seafood said:
>_> Oh. Well, I hope my explanation blurb thing helps so that I'm not just blatantly giving the solution without providing any real understanding.
PeterO is correct. The Physics Forums rules do not permit a member to post the solution to another member's problem.
 
  • #7
OK, I attached the rest of my work..
 

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  • #8
iRaid said:
OK, I attached the rest of my work..

In your 5th line, when you took out a factor of sin(theta) in the numerator, it was not a common factor, as it was in the denominator of a couple of the terms.
 
  • #9
PeterO said:
In your 5th line, when you took out a factor of sin(theta) in the numerator, it was not a common factor, as it was in the denominator of a couple of the terms.

I see now thanks, I got it. Stupid mistakes.
 

FAQ: Pretty difficult trig proof (identity)

1. What is a trigonometric identity?

A trigonometric identity is a mathematical equation that shows the relationship between different trigonometric functions. It is true for all values of the variables involved.

2. Why is this proof considered "pretty difficult"?

This proof may be considered difficult because it involves manipulating and simplifying complex trigonometric expressions using various identities and properties.

3. What are some common strategies for solving difficult trigonometric proofs?

Some common strategies for solving difficult trigonometric proofs include breaking down the problem into smaller parts, using identities and properties to simplify expressions, and substituting known values for variables.

4. How can I check if my solution to a trigonometric proof is correct?

You can check if your solution is correct by plugging in the values from the original equations into your solution and verifying that they are equal. You can also use a graphing calculator or trigonometric tables to check the values of the functions involved.

5. Can this proof be solved using a different approach?

Yes, there are often multiple ways to solve a trigonometric proof. You may choose to use different identities or properties, or approach the problem from a different angle. It is important to find a method that works best for you.

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