Summation of Finite trigonometrical series Let $O$ be any point on the circumference of a circle circumscribing a regular polygon $A_1,A_2,A_3.., A_{2n+1}$ such that $O$ lies on the arc $ A_1A_{2n+1} $. Show that $OA_1+OA_3+...OA_{2n+1}=OA_2+OA_4...+OA_{2n} $.

I tried by find the length of the sides from the angles $OPA_1 , OPA_2 , OPA_3 ... , OPA_{2n+1}$ Where i considered $P$ to be the circle with ,$OPA_1 = \theta$ and the radius of this circle to be $r$ .
Solving further i got struck and got a different answer. Can you pls help me solve this question in this method, If you have some other approach to this question then please share it.
Thank You
 A: The key formula that we need can be found HERE:
$$\sum_{k=0}^n\sin(\varphi + k\alpha)=\frac{\sin\frac{(n+1)\alpha}{2}\sin(\varphi+\frac{n\alpha}2)}{\sin\frac\alpha2}$$
If you apply the following identity:
$$\sin\alpha\sin\beta=\frac12(cos(\alpha-\beta)-cos(\alpha+\beta))$$
...you get:
$$\sum_{k=0}^n\sin(\varphi + k\alpha)=\frac{\cos(\frac\alpha2-\varphi)-\cos(\frac{2n+1}{2}\alpha+\varphi)}{2\sin\frac\alpha2}\tag{1}$$
Now, denote the center of circumscribed circle with $C$. Introduce: 
$$\angle OCA_{2n+1}=-2\varphi$$
$$\angle A_kCA_{2n+1}=k\alpha$$
$$\alpha=\frac{2\pi}{2n+1}\tag{2}$$
You can easily show that:
$$OA_k=2R\sin\frac{\angle A_kCA_{2n+1}-\angle OCA_{2n+1}}2=2R\sin(k\frac\alpha2+\varphi)$$
Now calculate two sums:
$$OA_1+OA_3+...OA_{2n+1}=2R\sin(\frac\alpha2+\varphi)+2R\sin(\frac{3\alpha}2+\varphi)+...+2R\sin(\frac{(2n+1)\alpha}2+\varphi)\tag{3}$$
$$OA_2+OA_4+...OA_{2n}=2R\sin(\alpha+\varphi)+2R\sin(2\alpha+\varphi)+...+2R\sin(n\alpha+\varphi)\tag{4}$$
Now apply formula (1) on (3) and (4) and simplify by taking into account (2). It should not be too complicated to show that (3) and (4) are actually equal.
