How to get from tan(a)/tan(b) to sin(a)/sin(b) or something simmilar

[Philip Robotic]

Homework Statement

In this task I have $$\frac{tana}{tanb}=c$$ (I know c)

In this task I am looking for $$\frac{sina}{sinb}=?$$

2. The attempt at a solution
I tried transforming the first equation using trig identities. I got this far and got stuck:
$$\left(\frac{tana}{tanb}\right)^{2}=\frac{(sina+sinb)\cdot (sina-sinb)}{sin^{2}b\cdot cos^{2}a}+1$$
Is there a trick to get from this to the fraction I am looking for?

(Please note this is a part of a bigger task and I don't know if it is even solvable).

 

[kuruman]

This is an approximation, not an equality. It is valid for small angles $a$ and $b$ usually less than 0.1 radians. The smaller the angles the better the approximation. Do some calculations using small values for $a$ and $b$, say $a=0.1$, $b=0.05$ and you will see what I mean.

More generally, the small angle approximation is $\tan a\approx\sin a\approx a$.

 

[epenguin]

Has that speck of dust surrounded my brain again?

Draw a picture. The ratio of two tangents is the ratio of two heights, both divided by the same base length which cancels out.

Ratio of two sines is ratio of the same two heights, both divided by the same hypotenuse which cancels out.

So isn't the ratio of two sines the same as the ratio of the corresponding two tangents, or am I having a bad day with this flu?

 

[kuruman]

Counterexample
$$\frac{\sin 50^o}{\sin 60^o}=0.884;~~\frac{\tan 50^o}{\tan 60^o}=0.688$$

I hope you beat the flu .

 

[Philip Robotic]

Referring to kuruman's first post, unfortunately the angles in this task are rather large, between 30 to 80 degrees (approximately). But they are smaller than 90 deg.

 

[kuruman]

Referring to kuruman's first post, unfortunately the angles in this task are rather large, between 30 to 80 degrees (approximately). But they are smaller than 90 deg.

Then the equality does not hold. See post #4.

 

[Philip Robotic]

I just spotted I made a mistake in my attempt at solution in post #1. It is correct now (but still unsolved ). I also graphed both of the functions and it seems that they nearly perfectly match when the bottom angle (b) is negative and the top stays positive. You can check it here:

https://www.desmos.com/calculator/al3zpm4jmd

 

[George Jones]

What does $\left( 1 + \tan^2 a \right)$ equal?

 

[epenguin]

Counterexample
$$\frac{\sin 50^o}{\sin 60^o}=0.884;~~\frac{\tan 50^o}{\tan 60^o}=0.688$$

ah yes should always try an example first.

Has that speck of dust surrounded my brain again?

Draw a picture. The ratio of two tangents is the ratio of two heights, both divided by the same base length which cancels out.

Ratio of two sines is ratio of the same two heights, both divided by NOT the same hypotenuse which NOTcancels out.

So isn't the ratio of two sines the same as the ratio of the corresponding two tangents, or am I having a bad day with this flu?

I hope you beat the flu .

Thank you I was worried for half an hour, but it looks ^^^ like there is no permanent impairment after all. Sorry for irrelevant interruption.

 

[Philip Robotic]

What does $\left( 1 + \tan^2 a \right)$ equal?

If I am not mistaken that is $\frac{1}{cos^{2}a}$ Thus I could cancel out the $cos^2a$ at the bottom of my equation and then I would have$$ \frac{(sina+sinb)\cdot (sina-sinb)}{sin^{2}b}+tan^2a\cdot\frac{(sina+sinb)\cdot (sina-sinb)}{sin^{2}b}+1$$ But I don't know what next should I do...
$tan^2a$ is not known alone, right?

 

[DaveE]

Really not clear what the problem is asking for. What form of "?" are they after?
For example, since tan(x) = sin(x)/cos(x) ⇒ sin(a)/sin(b) = c*cos(b)/cos(a) is this your answer? I have no idea.

 

[kuruman]

Really not clear what the problem is asking for. What form of "?" are they after?
For example, since tan(x) = sin(x)/cos(x) ⇒ sin(a)/sin(b) = c*cos(b)/cos(a) is this your answer? I have no idea.

As I understand it, the original post has been edited to ask the question, "Knowing the ratio of tangents, find the ratio of sines."

 

[Philip Robotic]

As I understand it, the original post has been edited to ask the question, "Knowing the ratio of tangents, find the ratio of sines."

Yes, that is it. Please note this is just a part of a bigger problem and that is as far as I got. I managed to find the ratio of tangents, but I need to have a ratio of sines.

 

[PeroK]

Homework Statement

In this task I have $$\frac{tana}{tanb}=c$$ (I know c)

In this task I am looking for $$\frac{sina}{sinb}=?$$

2. The attempt at a solution
I tried transforming the first equation using trig identities. I got this far and got stuck:
$$\left(\frac{tana}{tanb}\right)^{2}=\frac{(sina+sinb)\cdot (sina-sinb)}{sin^{2}b\cdot cos^{2}a}+1$$
Is there a trick to get from this to the fraction I am looking for?

(Please note this is a part of a bigger task and I don't know if it is even solvable).

Proposition: If $\frac{\tan a}{\tan b} = c$, then $\frac{\sin a}{\sin b} = f(c)$, where $f(c)$ is independent of $a, b$.

Is that proposition true or false?

 

[epenguin]

I think #8 is the thing to follow. However I do not think there is an answer in terms of the ratio of tan's alone, I think you need this ratio plus $sin a$ or some other trigonometrical function of a or b.
For your larger problem do you really need this? It needs the ratio of sines as input parameter? Can you just let that be the input?

 

[DaveE]

You can derive c = [sin(a+b) + sin(a-b)]/[sin(a+b) - sin(a-b)]
However, I doubt that that is useful for you.

 

[Philip Robotic]

Proposition: If tanatanb=ctan⁡atan⁡b=c\frac{\tan a}{\tan b} = c, then sinasinb=f(c)sin⁡asin⁡b=f(c)\frac{\sin a}{\sin b} = f(c), where f(c)f(c)f(c) is independent of a,ba,ba, b.

Is that proposition true or false?

If I understand the question correctly, it would be good if it were true, because a general relationship between these two fractions could be derived. I am not sure if is true or false, but since you asked this question it is false, right?

I think #8 is the thing to follow. However I do not think there is an answer in terms of the ratio of tan's alone, I think you need this ratio plus $sin a$ or some other trigonometrical function of a or b.
For your larger problem do you really need this? It needs the ratio of sines as input parameter? Can you just let that be the input?

Unfortunately the answer they are asking for is that fraction of sines (the question is about the relative refractive index)

 

[PeroK]

If I understand the question correctly, it would be good if it were true, because a general relationship between these two fractions could be derived. I am not sure if is true or false, but since you asked this question it is false, right?

You could always unleash the power of the spreadsheet to resolve the issue!

 

[Philip Robotic]

You could always unleash the power of the spreadsheet to resolve the issue!

I tried now it and unfortunately I can only find approximations with spreadsheet (referring back to post #4). The software didn't find a function that matches both of the fractions well enough for the task. Seems that I will have to try a different approach on the "big" problem itself.

I would really like to thank all of you: PeroK, Kuruman, epenguin, DaveE and Geogre Jones for devoting your time to helping me I really appreciate your effort

 

[kuruman]

I tried now it and unfortunately I can only find approximations with spreadsheet (referring back to post #4). The software didn't find a function that matches both of the fractions well enough for the task. Seems that I will have to try a different approach on the "big" problem itself.

I would really like to thank all of you: PeroK, Kuruman, epenguin, DaveE and Geogre Jones for devoting your time to helping me I really appreciate your effort

Perhaps if you posted the entire "task", one of us might be able to see a different approach from yours that may result in a tractable solution.

 

[PeroK]

I tried now it and unfortunately I can only find approximations with spreadsheet (referring back to post #4). The software didn't find a function that matches both of the fractions well enough for the task. Seems that I will have to try a different approach on the "big" problem itself.

I would really like to thank all of you: PeroK, Kuruman, epenguin, DaveE and Geogre Jones for devoting your time to helping me I really appreciate your effort

You underestimate the power of calculations. All you need to do is find $a_1, b_1$ and $a_2, b_2$ such that $\frac{\tan a_1}{\tan b_1} = \frac{\tan a_2}{\tan b_2}$, yet $\frac{\sin a_1}{\sin b_1} \ne \frac{\sin a_2}{\sin b_2}$

 

[Philip Robotic]

Perhaps if you posted the entire "task", one of us might be able to see a different approach from yours that may result in a tractable solution.

The thing is I can't post the entire task. It is a part of an ongoing physics contest. But thanks a lot for asking

 

[Philip Robotic]

You underestimate the power of calculations. All you need to do is find $a_1, b_1$ and $a_2, b_2$ such that $\frac{\tan a_1}{\tan b_1} = \frac{\tan a_2}{\tan b_2}$, yet $\frac{\sin a_1}{\sin b_1} \ne \frac{\sin a_2}{\sin b_2}$

Still working on it, but that's a great tip I am still new to computer calculations

 

[George Jones]

You underestimate the power of calculations. All you need to do is find $a_1, b_1$ and $a_2, b_2$ such that $\frac{\tan a_1}{\tan b_1} = \frac{\tan a_2}{\tan b_2}$, yet $\frac{\sin a_1}{\sin b_1} \ne \frac{\sin a_2}{\sin b_2}$

Yes, it is very easy to find specific examples of this. I was trying to show this with

What does $\left( 1 + \tan^2 a \right)$ equal?

but it is much easier to just pick some specific values.

If I am not mistaken that is $\frac{1}{cos^{2}a}$

Yes, now multiply both sides by $\sin^2 a$.

Still working on it, but that's a great tip I am still new to computer calculations

You don't need a computer, just use well-known values.

 

[Ray Vickson]

Still working on it, but that's a great tip I am still new to computer calculations

From $\tan a /\tan b = c$ we have $\tan a = c\, \tan b$, or $\tan^2 a = c^2 \tan^2 b.$ This can be re-written in terms of sines, as
$$\frac{\sin^2 a}{1-\sin^2 a} = c^2 \frac{\sin^2 b}{1-\sin^2 b}.$$ Solving for $\sin^2 a$ gives
$$\sin^2 a = \frac{c^2 \sin^2 b}{1 - \sin^2 b + c^2 \sin^2 b}.$$
For any values of $b$ and $c$ you pick, you can determine $\sin b$, so you can determine the required value of $\sin^2 a$, and hence the value of $\pm \sin a.$ Alternately, you can write
$$\frac{\sin a}{\sin b} = \pm \frac{c}{\sqrt{1+(c^2-1) \sin^2 b }}.$$

 

[Philip Robotic]

Alright...
Again, thank you all very much for help I've really learned a lot in this discussion from you
and
I have finally managed to find an alternative solution that doesn't require converting a fraction of tangents into a fraction of sines, but still it was a lot of fun trying to investigate that matter