Finding the Solution to an Equation: R.S = 1/1+cosx

  • Thread starter Veronica_Oles
  • Start date
In summary: Not sure what you mean by that. If you mean starting with the thing to prove in the form f(x)=g(x) then applying the same operation to each side of that equation, and repeating the process until you get something that is clearly true, no. That is not a valid way to prove the original statement. The implication is in the wrong direction. It is saying "if the thing to be proved is true then ..." E.g. if you simply multiply both sides by zero you will get a true statement, but you have proved nothing.However, that process can be used to help you find a proof. Having reached a statement that is clearly true, you can then see if all
  • #1
Veronica_Oles
142
3

Homework Statement


1/1+cosx = csc^2x - cscxcotx

Homework Equations

The Attempt at a Solution


R.S= csc^2x - cscxcotx
= (1/sin^2x) - (1/sinx * cosx/sinx)
= (1/sin^2x) - (cosx/sinx^2x)
= 1-cosx / sin^2x
= 1 - cosx / 1 - cos^2x
= 1 / 1 - cos^2x

It does not match the left side and I am unsure of what I did incorrectly.
 
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  • #2
Veronica_Oles said:
= 1 - cosx / 1 - cos^2x
= 1 / 1 - cos^2x
perhaps a simple typo error

see for yourself /check
 
  • #3
drvrm said:
perhaps a simple typo error

see for yourself /check
Where in the left side? Or in my actual work?
 
  • #4
Veronica_Oles said:
Where in the left side? Or in my actual work?

you have on the left side
1/(1+cosx)
and on right side you got ( 1-cosx) / (1-cos^2x);
the denominator can be written as (1+cosx).(1- cosx) using identity a^2- b^2 = (a+b).(a-b)
so the numerator gets canceled and you get 1/(1+cosx)
am i right?
 
  • #5
Veronica_Oles said:

Homework Statement


1/1+cosx = csc^2x - cscxcotx

Homework Equations

The Attempt at a Solution


R.S= csc^2x - cscxcotx
= (1/sin^2x) - (1/sinx * cosx/sinx)
= (1/sin^2x) - (cosx/sinx^2x)
= 1-cosx / sin^2x
= 1 - cosx / 1 - cos^2x
= 1 / 1 - cos^2x

It does not match the left side and I am unsure of what I did incorrectly.

This step here is the problem:

How did you go from

##=\frac{1-cos(x)}{1-cos^2(x)}## to

##=\frac{1}{1-cos^2(x)}## ?
 
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Likes drvrm
  • #6
SteamKing said:
This step here is the problem:

How did you go from

##=\frac{1-cos(x)}{1-cos^2(x)}## to

##=\frac{1}{1-cos^2(x)}## ?
I got rid of one of the cos from numerator then got rid of one from denominator.
 
  • #7
Veronica_Oles said:
I got rid of one of the cos from numerator then got rid of one from denominator.
You certainly did not do that since the denominator did not change. But even if you had it would have been invalid.
In a fraction, you cannot add something to the numerator and balance that by adding the same thing to the denominator. All you can do is multiply (or divide) the numerator as a whole by something and do exactly the same to the denominator. This is what drvrm did in post #4. Even then, you do need to be a bit careful. If the thing you multiply or divide by might be zero under some circumstances then you should exclude that by writing something like "when ... is not zero then..."
 
  • #8
haruspex said:
You certainly did not do that since the denominator did not change. But even if you had it would have been invalid.
In a fraction, you cannot add something to the numerator and balance that by adding the same thing to the denominator. All you can do is multiply (or divide) the numerator as a whole by something and do exactly the same to the denominator. This is what drvrm did in post #4. Even then, you do need to be a bit careful. If the thing you multiply or divide by might be zero under some circumstances then you should exclude that by writing something like "when ... is not zero then..."
Okay thank you. When doing trig identities you are able to work on both sides correct?
 
  • #9
Veronica_Oles said:
Okay thank you. When doing trig identities you are able to work on both sides correct?
Not sure what you mean by that. If you mean starting with the thing to prove in the form f(x)=g(x) then applying the same operation to each side of that equation, and repeating the process until you get something that is clearly true, no. That is not a valid way to prove the original statement. The implication is in the wrong direction. It is saying "if the thing to be proved is true then ..." E.g. if you simply multiply both sides by zero you will get a true statement, but you have proved nothing.
However, that process can be used to help you find a proof. Having reached a statement that is clearly true, you can then see if all the steps you took are reversible, and hence obtain the thing to be proved.
 

1. How do you solve for x in the equation R.S = 1/1+cosx?

To solve for x in this equation, you first need to isolate the variable on one side of the equation. You can do this by multiplying both sides by 1+cosx. This will give you R.S(1+cosx) = 1. Then, you can divide both sides by R.S to get 1+cosx = 1/R.S. Finally, you can subtract 1 from both sides to get cosx = 1/R.S - 1. From there, you can use trigonometric identities and inverse functions to solve for x.

2. What is the importance of finding the solution to this equation?

Finding the solution to this equation is important because it allows you to determine the value of x that satisfies the equation. This can help you understand the relationship between the variables and make predictions based on the given information.

3. What are the possible solutions for x in this equation?

The possible solutions for x in this equation will depend on the values of R and S. If R and S are both positive, there will be two solutions for x (one in the first quadrant and one in the second quadrant). If R is negative and S is positive, there will be one solution in the fourth quadrant. If R is positive and S is negative, there will be one solution in the third quadrant. And if R and S are both negative, there will be two solutions (one in the third quadrant and one in the fourth quadrant).

4. Can this equation have no solution?

Yes, it is possible for this equation to have no solution. This can occur if the value of 1+cosx is greater than 1/R.S, making it impossible to solve for x. It can also occur if the values of R and S are such that the equation has no real solutions.

5. How can this equation be applied in real-life situations?

This equation can be applied in various situations, such as in physics and engineering, to solve for unknown quantities or to understand the relationship between different variables. For example, it can be used to calculate the force required to move an object a certain distance or to determine the optimal angle for a projectile to be launched at.

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