Why Does Pi Equal 180 Degrees and 2 Pi Equal 360 Degrees?

[vee6]

I know a Pi is 3.14159265359 or 22/7.

Or Pi is a ration between a circle's circumference and its diameter.

Why Pi = 180 deg and 2 pi = 360 deg?

Please explain.

 

[Simon Bridge]

$\pi$

22/7 = 3.1429 i.e. it is not pi.
similarly pi is not 180 degrees.

Your second one is correct - pi is "Pythagoras' number" - so it carries the first letter of his name.
It is defined as the ratio of the circumference to the diameter of a circle.

22/7 is an old lower-school approximation for pi from before calculators were common - it is not used any more.

For a circle of radius R, an arclength of R subtends an angle, at the center, of 1 radien. This is the definition of a radien. Since there are 2pi radiuses in a full circumference, which is also 360 degrees, then one-pi radien must correspond to 180 degrees.

Note: 360 degrees to a circle means that the circumference of your protractor has to be 360 units. This means that the radius has to be an irrational number ... which can make building a very accurate one annoying. The bonus is that the numbers for common angles come out nice.

But if you choose radiens for your angle-measurements, it means that the radius is 1 unit long - which is easy to build. As a bonus there are lots of other simplifications as well - at the expense of having to keep that pesky pi in your calculations when you have simple angles. That turns out to be fine for scientific work since it often cancels out with a pi someplace else.

 

[lurflurf]

So in angle measure units

pi radians=180 degrees

or some people take degrees to be the constant

degrees=pi/180

then it is true that

pi=180 degees

 

[vee6]

What is the radian and pi relationship?

Pi is a pi.

Radian is a radian.

Is pi a relative?

 

[Simon Bridge]

What is the radian and pi relationship?

Pi is a pi.

Radian is a radian.

Is pi a relative?

I gave you the answer to that in post #2.

A pi is not a radian ... a radian is an angle.

pi is just a number - you can have pi amount of anything you like.

You can have pi radians just like you can have pi degrees and pi centimeters and pi anything.
Because of the way radians are defined, an angle of pi radians happens to be the same as an angle of 180 degrees.

The radian is defined to be the angle that gives an arc-length equal to the radius of the circle.
You work it out.

If you make a wheel that has a radius of 1m, and you roll it along the ground without slipping, and you stop when the wheel has turned 180 degrees ... how many meters along the ground did the wheel move?

 

[jonax101]

Spoiler

http://goo.gl/VhPTI meters.

 

[Simon Bridge]

Hi jonax101, welcome to PF.
There's also the sorority from Revenge of the Nerds.

 

[vee6]

So in angle measure units

pi radians=180 degrees

or some people take degrees to be the constant

degrees=pi/180

then it is true that

pi=180 degees

Which one is the constant?

Degrees or radians?

Your above statement seems the radians is the constants, not the degrees.

 

[vee6]

There is a method to find the arc length of circle by integration.

Why need a pi to find the circumference of a circle?

 

[Simon Bridge]

@vee6:
The questions you are asking are very basic - what level of education do you have?

Neither degrees nor radians are "constant". Radians are just more useful in some situations than in others.
You can define units to be anything you like ... I could, for example, make the "unit" for angle the circumference of the circle ... then all angles will be less than 1.

There is no "why" for needing pi to find the circumference of a circle - it just is: there is no other way it could be and still be a circle. Just like the the square on the hypotenuse is always the sum of the squares of the other two sides or 1+1=2.

Put another way: if you can find the circumference of a circle without pi, and prove it, then you will become very famous.

However:
You do not seem to be paying attention to the answers you are being given: you keep repeating questions that have already been answered. Why should anyone bother answering you when you don't learn?

 

[micromass]

There is a method to find the arc length of circle by integration.

Right, and if you follow that method, then $\pi$ will appear.

 

[marcusl]

So in angle measure units

pi radians=180 degrees

or some people take degrees to be the constant

degrees=pi/180

then it is true that

pi=180 degees

Not quite, you are being too casual with your numbers and units. The first line is correct. The second should be
1 degree = pi/180 radians.
The third line is then identical to the first.

As Simon Bridge already noted, there is no meaning to "degrees is the constant".

 

[vee6]

Right, and if you follow that method, then $\pi$ will appear.

How come?

Please write an example.

 

[HallsofIvy]

Which one is the constant?

Degrees or radians?

Your above statement seems the radians is the constants, not the degrees.

In other words you do not know what the word "constant" means. \pi is a number, about 3.1415962... (it is irrational and so cannot be written as a decimal number with a finite number of digits). That has nothing to do with "radians" or "degrees".

One application of \pi (admittedly the most common) is in circle measurement. If you use "radians" to measure angles, then a "straight angle" would be measured as \pi radians. If you used degrees, it would be 180 degrees. It is only in that sense that "\pi radians is the same as 180 degrees". Asking "Which is the constant? Degrees or radians?" is like, after being told that 1 meter is (approximately) the same as 39 inches, asking "which is the constant, meters or inches?"

The word "constant" can only be applied to numbers or functions of numbers, not to "units" such as radians and degrees (or meters and inches).

 

[vee6]

Right, and if you follow that method, then $\pi$ will appear.

Please write me an example, the integration to find the circle circumference that will makes the pi appears.

 

[micromass]

The circle is given by $(\cos(t),\sin(t))$. Then

\int_0^{2\pi} \sqrt{ x^\prime(t)^2 + y^\prime(t)^2 }dt = \int_0^{2\pi}\sqrt{\sin^2(t) + \cos^2(t)}dt = \int_0^{2\pi} dt = 2\pi

 

[Simon Bridge]

Please write me an example, the integration to find the circle circumference that will makes the pi appears.

Is it your contention that it is possible to use integration to find the circle circumference where pi does not appear?
In that case it is up to you to demonstrate it.
Can you produce an example of what you mean by an integration method that does not make pi appear?
If so then please do so - if not then, where did you get this idea from?

In a trig function like sin(t) that "t" is an angle ... so the "pi" is in the definition of the angle: just like has been explained to you repeatedly in the previous answers. You don't seem to believe the previous answers so there is no reason to think you will believe anything else we can tell you.

It will be far more useful for us to help you with these confusions if you will show us your understanding rather than us just telling you what is true.

 

[vee6]

The circle is given by $(\cos(t),\sin(t))$. Then

\int_0^{2\pi} \sqrt{ x^\prime(t)^2 + y^\prime(t)^2 }dt = \int_0^{2\pi}\sqrt{\sin^2(t) + \cos^2(t)}dt = \int_0^{2\pi} dt = 2\pi

Why use sin t and cos t instead y = f(x)?

The above pi already appears (tex]\int_0^{2\pi}[/tex]) before it appears (2pi).

Wrong if you say "Right, and if you follow that method, then $\pi$ will appear."

 

[micromass]

Why use sin t and cos t instead y = f(x)?

The above pi already appears (0, 2pi) before it appears (2pi).

Wrong if you say "Right, and if you follow that method, then $\pi$ will appear."

I don't see the objection, but if you want another method:

Take $f(x) = \sqrt{1 - x^2}$, then $f^\prime(x) = \frac{-x}{\sqrt{1-x^2}}$. Thus

\int_{-1}^1 \sqrt{1+ (f^\prime(x))^2}dx = \int_{-1}^1 \frac{1}{\sqrt{1 - x^2}}dx = arcsin(1) - arcsin(-1) =\pi

 

[Simon Bridge]

Why use sin t and cos t instead y = f(x)?

Having fun?
Which f(x) did you have in mind?

@Micromass: I don't think OP should get away with these vague statements: every example you try will have further objections:- nobody learns until they start doing their own work.

 

[vee6]

Still don't get the answer.

 

[micromass]

Still don't get the answer.

Can you say a bit more about what you don't understand and why not?

 

[Simon Bridge]

Still don't get the answer.

Then there is a communication problem - part of the problem is that you don't appear to want to say very much. For example: in the previous replies you have been asked a bunch of questions: you have yet to reply to any of them. These questions are not rhetorical - they are there to guide you (and us) to getting an answer you understand.

Please try to describe where we lose you - what is it you don't understand.
Don't worry about using the proper words or being clear - we get that you don't understand. We do understand ... we are used to people getting confused... and we are used to working through the confusion. Don't worry about looking silly: we've all been where you are - relax and gve it your best shot.

 

[SW VandeCarr]

Maybe a simpler approach. You accept that the circumference of a circle is \pi times the diameter. The radius is one half the diameter. Therefore the circumference is 2\pi times the radius. We call the arc length equal to the radius and the angle associated with it a "radian". Therefore there are 2\pi radians in the full arc length of the circle (equal to the circumference) and \pi radians in a the arc length of a half circle.

 

[Simon Bridge]

I have a feeling the persistent confusion lies in the way so many people think there is something mystical about pi.

 

[willem2]

I don't see the objection, but if you want another method:

Take $f(x) = \sqrt{1 - x^2}$, then $f^\prime(x) = \frac{-x}{\sqrt{1-x^2}}$. Thus

\int_{-1}^1 \sqrt{1+ (f^\prime(x))^2}dx = \int_{-1}^1 \frac{1}{\sqrt{1 - x^2}}dx = arcsin(1) - arcsin(-1) =\pi

If the first proof was cheating, this one is too.

you use arcsin (1) = pi/2, which implies that a 360 degree angle is 2pi, which is what you weren't allowed to use in the first proof.

 

[Simon Bridge]

you use arcsin (1) = pi/2, which implies that a 360 degree angle is 2pi, which is what you weren't allowed to use in the first proof.

Can you come up with one that isn't "cheating"?

I still thisnk the onus is on OP to come up with the method that doesn't have pi drop out somewhere.

 

[CarsonAdams]

This may help.

I think the issue here goes back to the fact that pi is a ratio of circumference to diameter. This seems fine, we're trying to see how many times your diameter will go around your circle, and that number is 3.14... times. Fine. Good. But wait, we always use the radius, not the diameter for calculations. So wouldn't a ratio of the circumference to the radius be more useful? Well that's another story, but it explains why RADIANS are so confusing for some.

You see, when we say that 2pi is a full circle, or 360 degrees, it seems odd. Why does it take 2pi radians (a type of angle measurement just like degrees but arguably scaled down) to go all the way around the circle?

The answer has a lot to do with the fact that a RADIan comes from the RADIus. That's where the word radian comes from, because when you've gone 1 radius length around your circles perimeter, the angle will be 1 radian. when you've gone pi radius lengths around your circle, its no surprise that the angle you get is pi radians, which will be a flat line.

This is, again, because if it takes pi number of diameter lengths to go around a circle, it will take 2pi radius lengths to go around the circle. This is why 2pi is 360 degrees.

As far as generating the value pi is concerned, the oldest way that requires no self-reliant definition is the method used by Archimedes. While granted this is basically early integration, Archimedes was simply using geometry. If you take a hexagon, and look at its perimeter to diameter ratio, you'll get 3. If you look at higher and higher numbered polygons, like a 17-gon, that ratio of circumference to diameter will slowly reach 3.14... which is exactly what you'd expect since a circle is just an infinitely sided polygon. Its a bit more complicated since he used both inscribed and circumscribed polygons to get an even better limit, but you get the gist. This isn't a great way to calculate pi, but Archimedes did it up to a 96-gon, and you can technically continue on this path infinitely to get pi definitionally. This was the first fully recorded estimation of pi in the western world to such an accuracy.

Didn't cheat did I?

 

[Simon Bridge]

Yep - the choice of the radius instead of the diameter as the unit in the unit circle is to do with how we draw circles.

All measurements involve comparing with some standard unit.
Distance requires a unit length, area requires a unit square (made from two unit lengths) and a unit volume is a cube whose sides are unit length and so on.

To describe the concept of an angle, you need a unit <shape> and the natural shape to use is a circle.
This is why we label angles with an arc.

A unit circle could have the radius, diameter or circumference to be equal to 1. Pick one ;)

The choice of unit determines how you measure angle.

Since you build circles by fixing the center and specifying the radius, the radius makes the most natural unit. So the unit circle is a circle with a radius which has unit length.

The size of the angle becomes the distance around the circumference of a unit circle inside the angle limits.
It's that simple.

But that's not the only way to build circles ... the other main way is to start out with a fixed length of something flexible, like a strip of metal, and bend it into a circle. This makes the circumference = one unit length.

Define the size of the angle the same way: the distance around the circumference of a unit circle, then all angles will have size less than one. To make arithmetic easier, we divide the circumference into 360 sub-units called "degrees". The main advantage of this method is that we don't get this pesky pi units in the angle (it's hidden in the diameter).

We could use the diameter as our unit ... that would make pi units in a full circle. Nothing wrong with that.

However, OP appears to have asserted that you can find the circumference from the diameter without using pi, vaguely refers to "integration" as the method, and then challenges everyone else to show that it is not possible.
Tough! I say it is up to OP to demonstrate this claim. The rest has been explained ad infinitum now ... it is up to OP to explain what is not adequate about the information supplied.

 

[middleCmusic]

Can you come up with one that isn't "cheating"?

I still thisnk the onus is on OP to come up with the method that doesn't have pi drop out somewhere.

I think your method is fair, as it doesn't use the numerical expression for pi - rather it just uses relationships between [the ratio of a circle's circumference and diameter] and right triangles. (The former of which happens to be pi).

If OP wants a numerical expression for pi - one that generates it - perhaps an infinite series is what OP wants (although I feel from the responses that OP may be missing some of the prerequisite material, but perhaps it's just a language barrier).

How about something like this:

\pi = 4(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \cdots)

I don't know the derivation though unfortunately.

EDIT: Thanks to Curious3141 for pointing out the missing ellipsis.

 

[Curious3141]

I think your method is fair, as it doesn't use the numerical expression for pi - rather it just uses relationships between [the ratio of a circle's circumference and diameter] and right triangles. (The former of which happens to be pi).

If OP wants a numerical expression for pi - one that generates it - perhaps an infinite series is what OP wants (although I feel from the responses that OP may be missing some of the prerequisite material, but perhaps it's just a language barrier).

How about something like this:

\pi = 4(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9})

I don't know the derivation though unfortunately.

Perhaps you meant this: \pi = 4(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - ...)

That ellipsis ('...') is all important.

Alternatively, you could've stated: \pi \approx 4(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9}), but the approximation is really quite mediocre with so few terms.

 

[middleCmusic]

Perhaps you meant this: \pi = 4(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - ...)

That ellipsis ('...') is all important.

Alternatively, you could've stated: \pi \approx 4(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9}), but the approximation is really quite mediocre with so few terms.

Oops! What an embarrassing mistake on my part. Fixed my post.

 

[micromass]

We have asked the OP repeatedly to clarify his ideas. He never clearly answered. And now he seems to be gone from this thread. Time to lock.