Prove: sinØ/(1-cosØ) - (1+cosØ)/sinØ = 0

[melya18]

Prove: sinØ/(1-cosØ) - (1+cosØ)/sinØ = 0

Prove: sinØ/(1-cosØ) - (1+cosØ)/sinØ = 0

I believe I need to use this: sin²θ = 1-cos²θ

But, how do I do that?

The textbook does not have any example problems similar to this at all!

Thanks for any help.

 

[stapel]

Prove: sinØ/(1-cosØ) - (1+cosØ)/sinØ = 0

A good first step, when given an identity with fractions, is to convert to a common denominator. In this case, that will be sin(@) (1 - cos(@)). After simplifying each fraction, you should get:

. . . . .\(\displaystyle \dfrac{\sin^2\left(\theta\right)}{\sin\left(\theta\right) \left(1\, -\, \cos^2\left(\theta\right)\right)}\, -\, \dfrac{1\, -\, \cos^2\left(\theta\right)}{\sin\left(\theta\right) \left(1\, -\, \cos^2\left(\theta\right)\right)}\)

Then combine into one fraction. You should get:

. . . . .\(\displaystyle \dfrac{\sin^2\left(\theta\right)\, +\, \cos^2\left(\theta\right)\, -\, 1}{\sin\left(\theta\right) \left(1\, -\, \cos^2\left(\theta\right)\right)}\)

Apply the basic Pythagorean identity, and the result immediately follows.

 

[HallsofIvy]

Prove: sinØ/(1-cosØ) - (1+cosØ)/sinØ = 0

I believe I need to use this: sin²θ = 1-cos²θ

But, how do I do that?

The textbook does not have any example problems similar to this at all!

Thanks for any help.

In order that \(\displaystyle \frac{sin(\phi)}{1- cos(\phi)}- \frac{1+ cos(\phi)}{sin(\phi)}= 0\) it is necessary that
\(\displaystyle \frac{sin(\phi)}{1- cos(\phi)}= \frac{1+ cos(\phi)}{sin(\phi)}\). (I added \(\displaystyle \frac{1+ cos(\phi)}{sin(\phi)}\) to both sides.)

In order that that be true, we must have \(\displaystyle sin^2(\phi)= (1+ cos(\phi)(1- cos(\phi))= 1- cos^2(\phi)\). (I multiplied both sides by \(\displaystyle sin(\phi)(1- cos(\phi))\).)

And in order for that to be true, we must have \(\displaystyle sin^2(\phi)+ cos^2(\phi)= 1\), which we know is true. (I added \(\displaystyle cos^2(\phi)\) to both sides.)

Now, start from the statement we know to be true, \(\displaystyle sin^(\phi)+ cos^2(\phi)= 1\) go backward by doing the opposite of each step. The opposite of adding a number to both sides is subtracting that number from both sides and the opposite of multiply a number on each side is dividing each side by that number. You will have to be careful to exclude value sof x for which cannot do that for example, you cannot divide by 0.

 

[melya18]

OK- that makes a lot of sense! I appreciate the guidance!

So I have:

sin² (Ø) + cos²(Ø) - 1/sin(Ø)(1-cos²)(Ø)

Adding in the sin²θ = 1-cos²θ =

Numerator: (1-cos²θ) + cos²(Ø) - 1 = cancel each other out = 0

Denominator: sin(Ø)(1-cos²)(Ø) ------> A little lost here????

sin(Ø)(1-cos²)(Ø)(1-cos²)(Ø) = sin(Ø)(sin²θ)?? or sin(Ø)(1-cos²θ)?? What am I doing wrong?

A good first step, when given an identity with fractions, is to convert to a common denominator. In this case, that will be sin(@) (1 - cos(@)). After simplifying each fraction, you should get:

. . . . .\(\displaystyle \dfrac{\sin^2\left(\theta\right)}{\sin\left(\theta\right) \left(1\, -\, \cos^2\left(\theta\right)\right)}\, -\, \dfrac{1\, -\, \cos^2\left(\theta\right)}{\sin\left(\theta\right) \left(1\, -\, \cos^2\left(\theta\right)\right)}\)

Then combine into one fraction. You should get:

. . . . .\(\displaystyle \dfrac{\sin^2\left(\theta\right)\, +\, \cos^2\left(\theta\right)\, -\, 1}{\sin\left(\theta\right) \left(1\, -\, \cos^2\left(\theta\right)\right)}\)

Apply the basic Pythagorean identity, and the result immediately follows.

 

[Ishuda]

Prove: sinØ/(1-cosØ) - (1+cosØ)/sinØ = 0

I believe I need to use this: sin²θ = 1-cos²θ

But, how do I do that?

The textbook does not have any example problems similar to this at all!

Thanks for any help.

Just to be pedantic: Actually the statement is not true for all \(\displaystyle \phi\). However it is true for all \(\displaystyle \phi\) such that both sin(\(\displaystyle \phi) \ne 0\) and cos(\(\displaystyle \phi) \ne 1\), i.e. for all \(\displaystyle \phi \ne 2 n \pi\), n an integer.

How so some ever,
\(\displaystyle \lim_{\phi \to 2 n \pi} \frac{sin(\phi)}{1 - cos(\phi)} - \frac{1 + cos\phi)}{sin(\phi)} = 0 \) for all integers n.

 

[stapel]

In order that \(\displaystyle \,\dfrac{\sin(\phi)}{1\,-\, \cos(\phi)}\,-\, \dfrac{1\,+ \,\cos(\phi)}{\sin(\phi)}\,=\, 0,\,\) it is necessary that \(\displaystyle \,\dfrac{\sin(\phi)}{1\,- \,\cos(\phi)}\,=\, \dfrac{1\,+ \,\cos(\phi)}{\sin(\phi)}\)....

Actually, if one works on both sides of the equation, one is assuming that the equation is true to start with. Technically speaking, all you'll have done is prove that, if the equation was really an identity, that then one ends up with something else. But this doesn't prove that the equation was really an identity. This is why one, technically, must work with only one side of the equation at a time.

While this may be a "technicality", many instructors will grade according to the mathematical logic. Proving an equation to be an identity (that is, proving it always to be true) is, logically, a process distinct from that of solving an equation which one assumed to be true (for at least some value of the variable) in the first place.

 

[HallsofIvy]

Actually, if one works on both sides of the equation, one is assuming that the equation is true to start with. Technically speaking, all you'll have done is prove that, if the equation was really an identity, that then one ends up with something else. But this doesn't prove that the equation was really an identity. This is why one, technically, must work with only one side of the equation at a time.

While this may be a "technicality", many instructors will grade according to the mathematical logic. Proving an equation to be an identity (that is, proving it always to be true) is, logically, a process distinct from that of solving an equation which one assumed to be true (for at least some value of the variable) in the first place.

I don't consider that a "technicality and I never said that was a proof. I said that if the identity was true then by doing these steps you could get the true identity \(\displaystyle sin^2(\phi)+ cos^2(\phi)= 1\). I then told the OP could he could get a valid proof by (carefully) reversing those steps.

 

[stapel]

I don't consider that a "technicality"....

I'm sure that you understand that. But many students (at least in my, admittedly limited, experience) go astray in thinking that the same techniques can be used in both situations, and I wanted to be sure that the difference was clear. I should have made my intention more clear. No offense or aspersions intended. Sorry!

 

[Steven G]

OK- that makes a lot of sense! I appreciate the guidance!

So I have:

sin² (Ø) + cos²(Ø) - 1/sin(Ø)(1-cos²)(Ø)

Adding in the sin²θ = 1-cos²θ =

Numerator: (1-cos²θ) + cos²(Ø) - 1 = cancel each other out = 0

Denominator: sin(Ø)(1-cos²)(Ø) ------> A little lost here????

sin(Ø)(1-cos²)(Ø)(1-cos²)(Ø) = sin(Ø)(sin²θ)?? or sin(Ø)(1-cos²θ)?? What am I doing wrong?

Be careful when you write (1-cos²)(Ø). You mean to write 1-cos²(Ø).

When is a fraction equal to 0? Answer, when the numerator is 0 and the denominator is not. So you have 0 on top. There is no real need to clean up the denominator. Here is why. sinØ/(1-cosØ) - (1+cosØ)/sinØ = 0 is NOT always true. It is not true when the denominator of any one of the fractions is 0. Although it is really sloppy in my opinion to call this an identity since it is NOT always 0, we ignore that detail. This means that whatever you get for the denominator is ok unless you get identically 0.

Having said that sin(Ø)(1-cos²θ)=sin(θ)sin²(θ)=sin^3(θ)

 

[pvk21]

Why dont use cos@ = 1-2sin^2@ and cos@ = 2cos^2@-1. Its easier that way

 

[pvk21]

Its cos2@ not cos@ ..sorry for mistake

 

[pvk21]

Why dont use cos2@=1-2sin^2@ and cos2@=2cos^2@-1. Its easier that way.