This place has changed... and a math question with a little trigonometry :P

[Kerr]

Hi... this place has changed since I was last here (have just had a lot to do with my studies). So much that I forgot where to ask this (or maybe you can't, lol). Asking here as an excuse to get back, btw, I know that I should probably ask at a math forum or something.

Anyway, to the math question. It has to do with trigonometry and approximations. Let X be an angle. If X is small enough we can approximate sin(X) as tan(X). So... how small does X have to be for this to be a valid approximation?

Asking because I had this exercise to do and in the solution they used that approximation and got a completely different result then me. Only thing is, X was about 41-42 degrees. That seems a bit... large... for a small angle (which was why I didn't use that approximation).

Ok, so now that I have asked my math question, how is everyone doing?

Take care,
Kerr.

EDIT:

Ok, I found the right place. It was the "homework and tutoring" part of the forums. Was a bit confused by the new appearence so I missed it, lol. If anyone feel like moving this thread feel free to do so.

 

[Iti oj]

I didn't even know that sub forum.existed. btw welcome back

 

[Revoltingest]

Hi... this place has changed since I was last here (have just had a lot to do with my studies). So much that I forgot where to ask this (or maybe you can't, lol). Asking here as an excuse to get back, btw, I know that I should probably ask at a math forum or something.

Anyway, to the math question. It has to do with trigonometry and approximations. Let X be an angle. If X is small enough we can approximate sin(X) as tan(X). So... how small does X have to be for this to be a valid approximation?

Asking because I had this exercise to do and in the solution they used that approximation and got a completely different result then me. Only thing is, X was about 41-42 degrees. That seems a bit... large... for a small angle (which was why I didn't use that approximation).

Ok, so now that I have asked my math question, how is everyone doing?

Take care,
Kerr.

EDIT:

Ok, I found the right place. It was the "homework and tutoring" part of the forums. Was a bit confused by the new appearence so I missed it, lol. If anyone feel like moving this thread feel free to do so.

Welcome back!
I faced this very issue yesterday in calculating the thrust angle (which causes dog tracking) on one of my equipment trailers. Taking the arc sign of one dimension over another, I found 1.38 degrees. The tangent & sine of this angle were close enuf for my tolerances. It depends upon your application.

 

[Kerr]

Welcome back!

Thanks!

I faced this very issue yesterday in calculating the thrust angle (which causes dog tracking) on one of my equipment trailers. Taking the arc sign of one dimension over another, I found 1.38 degrees. The tangent & sine of this angle were close enuf for my tolerances. It depends upon your application.

It's about optics (and about me being rather picky).

The problem is to find the height of a barrel filled with water if we know that when the sun is 28 degrees over the horizon light no longer reaches the bottom (the refraction index of water is given as 1.33 and the diameter of the barrel is 3 meters). The solution got 3.38 meters as the result. I got approximately 5.42 meters. The main difference between our calculations seems to be the sin(X) = tan(X) approximation. I get why they want to do that approximation, it's simplifies a lot, I'm just not conviced it would give the right result. I mean, sin(41.6) = 0.66 while tan(41.6) = 0.89. Seems to different to me to be valid. But maybe I am missing something.

EDIT:

If I take the arc sine of tan(41.6) I get 62.6 degrees. If the approximation was good, shouldn't I get closer to 41.6 degrees?

 

[Ouroboros]

Anyway, to the math question. It has to do with trigonometry and approximations. Let X be an angle. If X is small enough we can approximate sin(X) as tan(X). So... how small does X have to be for this to be a valid approximation?

An approximation is always an approximation. There's no "valid" to it except how close you want your approximation to be.

Remember that tan(x) is sin(x)/cos(x). The reason you can use it for approximation for small x is because cos(x) gets closer to 1. In other words, how well the approximation works is in relation to how close cos(x) is to 1.

Asking because I had this exercise to do and in the solution they used that approximation and got a completely different result then me. Only thing is, X was about 41-42 degrees. That seems a bit... large... for a small angle (which was why I didn't use that approximation).

Yes, that sounds like a very high number.

 

[Ouroboros]

The problem is to find the height of a barrel filled with water if we know that when the sun is 28 degrees over the horizon light no longer reaches the bottom (the refraction index of water is given as 1.33 and the diameter of the barrel is 3 meters). The solution got 3.38 meters as the result. I got approximately 5.42 meters. The main difference between our calculations seems to be the sin(X) = tan(X) approximation. I get why they want to do that approximation, it's simplifies a lot, I'm just not conviced it would give the right result. I mean, sin(41.6) = 0.66 while tan(41.6) = 0.89. Seems to different to me to be valid. But maybe I am missing something.

Where did you get the 41.6º from? Without having all the information, you pay attention to the quadrants? It's not just that you suddenly move into the second or something, changing the signs of the trig functions.

EDIT:

If I take the arc sine of tan(41.6) I get 62.6 degrees. If the approximation was good, shouldn't I get closer to 41.6 degrees?

It was a couple of years now since I learned these things, so I'm not sure at the moment what goes wrong. The only thing I know is that these angels seem to be way high to be used in the tan/sin approximation.

 

[Kerr]

Where did you get the 41.6º from? Without having all the information, you pay attention to the quadrants? It's not just that you suddenly move into the second or something, changing the signs of the trig functions.

It comes from Snell's law. If we call the angle of incidence (the angle a light ray hit the water at, against the normal) for A and the angle of refraction (the angle of the bent light against the normal of the water) for B, we can get B as B = arcsin(sin(A)/n). A is 90 - 28 = 62 degrees. The refraction index of air is assumed to be 1 and the refraction index for water is assumed to be 1.33. If we put this in the equation we get that B = 41.6 degrees (if we round to one decimal).

It was a couple of years now since I learned these things, so I'm not sure at the moment what goes wrong. The only thing I know is that these angels seem to be way high to be used in the tan/sin approximation.

No worries, I tend to forget rather quickly when I don't actively use some knowledge xD. And you made a good point about my wording with "valid approximation". Should probably have said "good approximation" or "good enough approximation for my taste", lol.

 

[Ouroboros]

It comes from Snell's law. If we call the angle of incidence (the angle a light ray hit the water at, against the normal) for A and the angle of refraction (the angle of the bent light against the normal of the water) for B, we can get B as B = arcsin(sin(A)/n). A is 90 - 28 = 62 degrees. The refraction index of air is assumed to be 1 and the refraction index for water is assumed to be 1.33. If we put this in the equation we get that B = 41.6 degrees (if we round to one decimal).

Ok.

The 90-28 raises a flag for me. There's something going on here with the signs between sin and tan.

No worries, I tend to forget rather quickly when I don't actively use some knowledge xD. And you made a good point about my wording with "valid approximation". Should probably have said "good approximation" or "good enough approximation for my taste", lol.

I hear ya'. If I can remember where I put my glasses... I'm starting to looking for them while I'm wearing them. LOL!

 

[Kerr]

Ok.

The 90-28 raises a flag for me. There's something going on here with the signs between sin and tan.

Not sure it's the sign. Both sin and tan are positive for a 62 degree angle. I just don't think cos for that angle is close enough to 1 that the approximation yields good results. Or I have calculated it wrong and blame the approximation, that could happen xD. This is the solution I am suspect off:

I hear ya'. If I can remember where I put my glasses... I'm starting to looking for them while I'm wearing them. LOL!

Lol, that has never happened to me... yet... that I remember xD.

 

[Ouroboros]

Not sure it's the sign. Both sin and tan are positive for a 62 degree angle. I just don't think cos for that angle is close enough to 1 that the approximation yields good results. Or I have calculated it wrong and blame the approximation, that could happen xD. This is the solution I am suspect off:

View attachment 8943


Lol, that has never happened to me... yet... that I remember xD.

The theta angle looks like the 28 degrees, not the 62.

 

[Kerr]

Ok, I actually solved it! It wasn't the approximation, lol. It was that I had accidently replaced a - with a + at a convienient location xD. I still think it's a suspicious approximation, though, even if the result seems very close. I wonder... if the angle is close to 45 degrees, is that approximation "good enough" again or something? Or are they doing something else and I am too tired to see it and instead I just see a weird approximation?

 

[Kerr]

The theta angle looks like the 28 degrees, not the 62.

The theta angle is the angle of refraction and it wont be 28 or 62 degrees. It's the one that was calculated to 41.6 degrees. Sorry if the picture isn't clear, btw. I just print screened it onto this forum xD.

 

[Ouroboros]

The theta angle is the angle of refraction and it wont be 28 or 62 degrees. It's the one that was calculated to 41.6 degrees. Sorry if the picture isn't clear, btw. I just print screened it onto this forum xD.

Oh. Right. I had a *doh* moment.

 

[Kerr]

Oh. Right. I had a *doh* moment.

We both had xD.

(In my case it was the replacing - with +)