What is the proof for the rare trig identity with tan/tan = other/other?

[Blanchdog]

Homework Statement

Prove that $\frac{\text{tan}(x - y)}{\text{tan}(x + y)} = \frac{ \text{sin}(x)\text{cos}(x) - \text{sin}(y)\text{cos}(y)}{\text{sin}(x)\text{cos}(x) + \text{sin}(y)\text{cos}(y)}$

Relevant Equations

Other trig identities.

Came across this trig identity working another problem and I've never seen it before in my life. I don't need to prove it myself, necessarily, but I would really like to see a proof of it (my scouring of the internet has yielded no results). If someone more trigonometrically talented than myself would take a crack at it, or else guide me through at least the intuition behind it that would be wonderful. Thanks!

 

[.Scott]

I can prove it is not:

Since the ranges of sin and cos are -1 to +1, the range of the product cannot exceed -1 to +1. (In fact, it's -0.5 to +0.5).

Since we are adding two of these products, the range cannot exceed -2 to +2.

The range of the tan function is +/- infinity, far out-ranging the sum of those products.

Thus, $tan(x+y)$ cannot equal $sin(x)cos(x)+sin(y)cos(y)$.

Here is a specific counter-example:
$tan(0.7+0.8) = tan(1.5) = 14.1014$
$sin(0.7)cos(0.7)+sin(0.8)cos(0.8) = 0.4927 + 0.4998 = 0.9925$
14.1014 is not equal to 0.9925

 

[cbarker1]

do you know sum or difference identities for tangent ?

 

[.Scott]

do you know sum or difference identities for tangent ?

Look here.

 

[cbarker1]

Look here.

My question was meant for the person who posted the homework question, sorry for the confusion.

 

[Haborix]

I can prove it is not:

Since the ranges of sin and cos are -1 to +1, the range of the product cannot exceed -1 to +1. (In fact, it's -0.5 to +0.5).

Since we are adding two of these products, the range cannot exceed -2 to +2.

The range of the tan function is +/- infinity, far out-ranging the sum of those products.

Thus, $tan(x+y)$ cannot equal $sin(x)cos(x)+sin(y)cos(y)$.

Here is a specific counter-example:
$tan(0.7+0.8) = tan(1.5) = 14.1014$
$sin(0.7)cos(0.7)+sin(0.8)cos(0.8) = 0.4927 + 0.4998 = 0.9925$
14.1014 is not equal to 0.9925

I don't think this is a good argument. Consider the following:

$1=\frac{-1}{-1}=\frac{x^2}{x^2}$

Therefore this equality is not true because $x^2\geq 0$ and so can never equal $-1$. Showing two numerators or denominators are not equal doesn't prove the ratio doesn't hold.

 

[phinds]

I can prove it is not:

You are obviously correct, but the OP has now gone back and modified the original problem to be a different problem.

 

[phinds]

I don't think this is a good argument. Consider the following:

$1=\frac{-1}{-1}=\frac{x^2}{x^2}$

Therefore this equality is not true because $x^2\geq 0$ and so can never equal $-1$. Showing two numerators or denominators are not equal doesn't prove the ratio doesn't hold.

What ARE you talking about? I don't see how setting x**2 to -1 proves or disproves anything and that's all you've done. How does this have any bearing on the ORIGINAL problem, which is what @.Scott was addressing.

 

[Delta2]

@phinds what was the ORIGINAL problem?
@.Scott post #2 just proves that $\tan(x+y)\neq \sin x\cos x+\sin y\cos y$ but this is not what the current problem is asking about..

 

[phinds]

@phinds what was the ORIGINAL problem?
@.Scott post #2 just proves that $\tan(x+y)\neq \sin x\cos x+\sin y\cos y$ but this is not what the current problem is asking about..

What @.Scott posted and said cannot be true (because it can't). That was the original question. As I said, the OP went back and changed the problem after he had already gotten an answer. Very bad form here on PF to do that.

 

[Delta2]

What @.Scott posted and said cannot be true (because it can't). That was the original question. As I said, the OP went back and changed the problem after he had already gotten an answer. Very bad form here on PF to do that.

So the original problem was something like "Prove that $$\tan(x+y)=\sin x\cos x+\sin y \cos y$$"?

 

[.Scott]

Homework Statement:: Prove that $tan(x+y) = sin(x)cos(x) + sin(y)cos(y)$

Relevant Equations:: Other trig identities.

Came across this trig identity working another problem and I've never seen it before in my life. I don't need to prove it myself, necessarily, but I would really like to see a proof of it (my scouring of the internet has yielded no results). If someone more trigonometrically talented than myself would take a crack at it, or else guide me through at least the intuition behind it that would be wonderful. Thanks!

What I show above is post #1 as it was originally posted. It was what I responded to in post #2.

The following is no different than my post #2 above, but this time I have quoted what I am disproving (the original post #1).

A cursory inspection of this formula can demonstrate that it is false.
Looking at the right-hand side of the "equation", you have the sum of two sin,cos products. Sin and cos are in the range of -1 to 1, so the products will also be within the range of -1 to 1. The sum of two of them will be within the range of -2 to 2. Since this is inconsistent with the range of the left hand side, the equation is false.

I am pretty sure that none of the posters intended to challenge this proof.
If I am wrong, let me know.

 

[willem2]

Homework Statement:: Prove that $\frac{\text{tan}(x - y)}{\text{tan}(x + y)} = \frac{ \text{sin}(x)\text{cos}(x) - \text{sin}(y)\text{cos}(y)}{\text{sin}(x)\text{cos}(x) + \text{sin}(y)\text{cos}(y)}$

The right hand side is obviously sin(x-y)/sin(x+y), implying that cos(x+y) = cos(x-y), so this new version isn't true either.
if you look up tan(x-y) / tan(x+y) on wolfram alpha, you can actually find a trig identity that IS true. You can prove it by using the addition formula's for sin and cos, and tan(x) = sin(x)/cos(x).

 

[Delta2]

The right hand side is obviously sin(x-y)/sin(x+y), implying that cos(x+y) = cos(x-y), so this new version isn't true either.

Nope, I was also thinking along the same lines but I corrected myself just before posting. It is $$\sin(x-y)=\sin x\cos y-\sin y \cos x$$ and similar for $sin(x+y)$.

 

[pasmith]

I think the result follows straightforwardly from $$\frac{\tan(x-y)}{\tan(x+y)} = \frac{\sin(x-y)\cos(x+y)}{\cos(x-y)\sin(x+y)}.$$

 

[phinds]

Your original argument was:

Thus, $tan(x+y)$ cannot equal $sin(x)cos(x)+sin(y)cos(y)$.

Which I agree with.

NOW, with the tan/tan = other/other, your argument doesn't seem to me to hold. The absolute value of the tan & other is irrelevant, since it is a RATIO that is being equated.

 

[Haborix]

What I show above is post #1 as it was originally posted. It was what I responded to in post #2.

The following is no different than my post #2 above, but this time I have quoted what I am disproving (the original post #1).

A cursory inspection of this formula can demonstrate that it is false.
Looking at the right-hand side of the "equation", you have the sum of two sin,cos products. Sin and cos are in the range of -1 to 1, so the products will also be within the range of -1 to 1. The sum of two of them will be within the range of -2 to 2. Since this is inconsistent with the range of the left hand side, the equation is false.

I am pretty sure that none of the posters intended to challenge this proof.
If I am wrong, let me know.

Ah, thank you for the clarification. I had only been able to see the revised question when I read your first proof post.

 

[.Scott]

NOW, with the tan/tan = other/other, your argument doesn't seem to me to hold. The absolute value of the tan & other is irrelevant, since it is a RATIO that is being equated.

That's right. With the new equation, the numerator and denominator ranges can cross zero - so there is no cursory proof.