How to solve the equation?

[DimaMazin]

a = (Pi - 2)/4 + sin(a)

a is unknown angle(rad) in unit circle

a/2 is area of sector of angle a

(Pi-2)/8 is area of segment of angle a

[joigus]

This is a transcendental equation. You cannot solve but by means of approximate methods, like iterations that are known to converge to a solution, etc.

[John Cuthber]

I wonder if there's a geometrical solution.

[studiot]

Firstly @DimaMazin

 

Please number your equations for discussion in future.

Calling them 1, 2 and 3 in the order in which they appear in the OP,

This is not a valid question.

Equation 1 may be simply derived from equation 3 and is an identity so valid for all values of a, where a is measured in radians.

 

Since I can no longer use either of my maths computers here you will all have to put up with crappy english language.

 

The standard expression for the area, A, of a segment of a circle of radius R is 

A  =  R^2/2 (a - sina)

Since you say R = 1

We have

(Pi - 2) / 8  =  1/2 (a - sina)

Multiply through by 2

( Pi - 2 ) / 4  = a - sina

Which is your equation 1

 

Perhaps you are looking for an angle a where the area is the same for both sector and segment ?

 

Edited April 18, 2023 by studiot

[Genady]

a=1.226819...

[studiot]

33 minutes ago, Genady said:

a=1.226819...

How does that work out with Dima's equation2, given that the standard area for a sector is Ra ?

[Genady]

5 minutes ago, studiot said:

How does that work out with Dima's equation2, given that the standard area for a sector is Ra ?

Area of a sector with angle a in radians and unit radius is

a/(2p)*(p*r^2)=a/2

for any a.

[studiot]

37 minutes ago, Genady said:

Area of a sector with angle a in radians and unit radius is

a/(2p)*(p*r^2)=a/2

for any a.

Yes you are right, the formula is R^2a/2  = a/2 in this case, so it does fit  +1

[joigus]

Genady is right. It's not a transcendental equation. I made a mistake. Sorry. Somehow I thought I saw an "a" in the first term, which wasn't there.

[Genady]

5 hours ago, joigus said:

Genady is right. It's not a transcendental equation. I made a mistake. Sorry. Somehow I thought I saw an "a" in the first term, which wasn't there.

I think, you were right, and it is a transcendental equation. I've solved it with a very simple approximation procedure and Excel. Here it is:

f(x) = (Pi - 2)/4 + sin(x) - x

[joigus]

4 hours ago, Genady said:

I think, you were right, and it is a transcendental equation. I've solved it with a very simple approximation procedure and Excel. Here it is:

Yes, thank you. It is. Every time I look at this thread I look without looking, if you know what I mean.

a = const. + sin(a) is a transcendental equation. 

Doh!!

[DimaMazin]

I have got so complex equation for definition of sine of the angle a ,but I am not sure it is correct.

4sin² +4sin*cos-2sin-Pi*sin+Pi*cos-2cos=0

If we use Genady's definition then we can approximately check it.

I have used method of disproportionate division of segment of angle Pi/2 (area of wich is (Pi-2)/4) and second part of the sector of the angle Pi/2 ,area of wich is 1/2. It is when angle a and angle Pi/2 -a  disproportionately divide the parts. Then every of the 4 parts has the same disproportionate unknown(for 2 parts it is u and for 2 other parts it is -u).

It is not working for definition sine and cosine of 1 radian relative to Pi, therefore rather it is nonsense.

 

[DimaMazin]

On 4/18/2023 at 11:24 AM, John Cuthber said:

I wonder if there's a geometrical solution.

Yes. But we can solve simpler problem.       a=2sin(a)

((sin-P/4)*2-sin*(-cos))*2 = sin-2sin*(-cos)-4(sin-P/4)

sin = (2/7)P

a=(4/7)P

Area of segment of angle a  =  (1/7)P

Area of segment of angle (3/14)P = (1/14)P

Exuse me. I have mistaken again. I incorectly made the equation. In correct equation the variables annihilate.

Edited June 12, 2023 by DimaMazin

[DimaMazin]

I think the simplest problem, of the similars,  is   sector area = chord²

a/2 = 2-2cos(a)

But I don't know how to solve it.

 

[DimaMazin]

Let's consider speculation how to solve a=2sin(a). 

Sector of such angle is special. Any sector has two parts of its area. One part of its area=sin/2. And second part is segment area. Two parts of the sector are equal. Then let's consider else two sectors. Sector of  half of the angle a/2  and sector of angle (Pi-a)/2+a.  My speculative idea is: 

(area of segment of angle(Pi-a)/2+a) ÷ sin(of the angle)/2 = sin(a/2) ÷ (area of segment of angle a/2)

Can it be so?

 

[Genady]

a=1.8954942670...

[DimaMazin]

On 12/10/2024 at 5:09 PM, Genady said:

a=1.8954942670...

Thanks. My idea is wrong.