after trying so much time and effort I still can't understand what this is page says. I understand by very simple language, Can someone please explain it in plan English what this book is saying, what I could understand is that it is about some ratio between line segments of a polygon inscribed in circular segment, and at the end it says its relation with trig function, First i don't understand what does is say about polygon and also I cant get it how it is related to following trig equation. if someone can tell me in a easy to understand manner, may be I can understand it. Thanks.
In this proposition $I, 22$ from On The Sphere and Cylinder Archimedes shows, mostly by similar triangles, that$$\frac{BB'+CC'+...+LM}{AM}=\frac{A'B}{BA}$$A trigonometric equivalent can be derived if we first multiply through by $AM$, to get$$BB'+CC'+...+LM=AM\cdot \frac{A'B}{BA}$$Now if $n$ is the even number of equal arcs into which the arc $LL'$ of the segment has been divided, and we denote $\angle LOL'$ as $\theta$, then as @Intelligenti pauca notes$$\angle BOA=\frac{\theta}{n}$$
Letting diameter $AA'=1$, then in the figure below, since $\triangle BB'F$ is right and similar to $\triangle BDO$, and $BF=AA'=1$, we get$$BB'=\sin\frac{\theta}{n}$$
Next, $\angle COA=\frac{2\theta}{n}$, and since $\triangle CC'G$ is right and similar to $\triangle CEO$, and $CG=1$, then $$CC'=\sin\frac{2\theta}{n}$$and so on.
We end up, however, not with $n$ terms on the left side of the equation, as the posted text says, but $\frac{n}{2}$ terms, the last of which, the equivalent of $LM$, should be$$\frac{1}{2}\sin\frac{n\theta}{2n}$$
Turning to the right side of the equation, since $\angle AA'B=\frac{1}{2}\angle AOB$, and $\angle ABA'$ is right, we have$$\frac{A'B}{BA}=\cot\frac{\theta}{2n}$$in agreement with the text.
But lastly, since $L'H=2MO=\cos\frac{\theta}{2}$, and $2AM$ is here the versed sine of $\frac{\theta}{2}$, or $$1-\cos\frac{\theta}{2}$$ then$$AM=\frac{1-\cos\frac{\theta}{2}}{2}$$which again does not exactly agree with the posted text.
For a trigonometric equivalent of Archimedes' claim, then, I am left with$$\sin\frac{\theta}{n}+\sin\frac{2\theta}{n}+...+\sin\frac{\frac{n}{2}-1}{n}\theta+\frac{1}{2}\sin\frac{\frac{n}{2}\theta}{n}=\frac{1-\cos\frac{\theta}{2}}{2}\cot\frac{\theta}{2n}$$
I am afraid this may increase OP's difficulties as much as lessen them. But it may at least help to show some of the geometry-trigonometry connection. I would like to know the context of the question. Was the text mainly concerned with Archimedes or with trigonometry? If with Archimedes, and OP has gotten as far as Proposition I, 22, that proposition seems to require knowing only a little Euclidean geometry--similar triangles and theory of proportions,--and the trigonometry seems an unnecessary distraction. But like OP, I welcome further insights or corrections.
Correction: OP's posted text was from C. B. Boyer A History of Mathematics, pp. 145-6.
Adopting the suggestion of @Intelligenti pauca, that for Boyer $\angle\theta=AOL$ and the number of equal arcs is $2n$, then if the circle's radius is $1$, for $BB'+CC'+...KK'+LM$ we get on the left$$2\sin\frac{\theta}{n}+2\sin\frac{2\theta}{n}+...+2\sin\frac{(n-1)\theta}{n}+\sin\frac{n\theta}{n}$$
On the right, as before$$\frac{BA'}{BA}=\cot\frac{\theta}{2n}$$But since $AM$, now the versed sine of $\theta$, is $1-\cos\theta$, we get the equation$$2\sin\frac{\theta}{n}+2\sin\frac{2\theta}{n}+...+2\sin\frac{(n-1)\theta}{n}+\sin\frac{n\theta}{n}=(1-\cos\theta)\cot\frac{\theta}{2n}$$and division by $2$ gives Boyer's equation$$\sin\frac{\theta}{n}+\sin\frac{2\theta}{n}+...+\sin\frac{(n-1)\theta}{n}+\frac{1}{2}\sin\frac{n\theta}{n}=\frac{1-\cos\theta}{2}\cot\frac{\theta}{2n}$$