Is this proof for area of a circle correct? I want to prove that the area of a circle is $\pi r^2$.
so I calculated the area of a regular polygon with $n$ sides. It consists of $n$ triangles with height of $r \cos(\frac{\pi}{n})$ and base of $2r \sin(\frac{\pi}{n})$ so :
$$S=2r^2\cos\left(\frac{\pi}{n}\right)\sin\left(\frac{\pi}{n}\right)=\frac{1}{2}r^2\sin\left(\frac{2\pi}{n}\right)$$
$$A=nS$$
$S$ is the area of each triangle and $A$ is the area of the polygon.
and when $n\rightarrow \infty$ 
$$A=n\left(\frac{1}{2}r^2\sin\left(\frac{2\pi}{n}\right)\right)=n\left(\frac{1}{2}r^2\frac{2\pi}{n}\right)$$
$$A=\pi r^2$$
Is it correct ?
 A: In the 18th century Leonhard Euler would have said that if $n$ is an infinitely large integer, then $\sin\dfrac{2\pi}n = \dfrac{2\pi}n$.  Today we say that the limit as $n\to\infty$ of the ratio of one of those to the other is $1$.
I'd say the argument is on the right track but seems more complicated than it needs to be and seems to rely on less primitive concepts than what is needed.  I also have some qualms about nitpicking details.
One may consider the polygon as inscribed in the circle or circumscribed about the circle or some compromise between those extremes, and as far as I know it doesn't matter which you pick.
Suppose we consider it circumscribed rather than inscribed and we eschew trigonometric functions and argue as follows: the area of each triangle is 1/2 base times height.  The sum of the bases is the perimeter of the polygon, so the area of the polygon is 1/2 perimeter times "radius", where the "radius" is the distance from any vertex to the center.
By making $n$ large enough, we can make the perimeter of the polygon differ by as little as desired from the circumference of the circle, and we can make the area of the polygon differ by as little as desired from the area of the circle.  Thus 1/2 times the circumference of the circle times the radius can be made as close as desired to the area of the circle by making $n$ big enough.  But the circumference of the circle and the radius and the area of the circle do not depend on $n$.  Therefore 1/2 times the circumference times the radius must be exactly the area of the circle.
A: As pointed by Andre Nicolas the proof has a flaw when taking the limit as $n \to \infty$. A better proof which is based on the assumption

Circumference of a circle bears a constant ratio to its diameter and that ratio is denoted by $\pi$.

can be given in the following manner. Cut $2n$ sectors of equal area and arrange them like (sorry for the quality of the figure below, its drawn in haste)

If the number $n$ is large the width of the sectors will be very small so that the both radial lines of each sector will almost coincide and the curved portion on top (and bottom) of the sectors will becomes almost like a very small straight line. The figure will then transform into a rectangle whose height is $r$ the radius of circle. The width on the other hand will be half the circumference $(1/2)(2\pi r) = \pi r$. And area of rectangle is $\pi r^{2}$.
However it suffers from the same defect as the one you mention. It only avoids mention of trigonometric function like $\sin$. The basic fact which it assumes is that the ratio of chord of a circle and its corresponding arc tends to $1$ as the arclength (or chordlength) tends to $0$.
A proper proof of the area of circle is via integrals where we can show that the area of circle of radius $r$ is equal to $$A = 4\int_{0}^{r}\sqrt{r^{2} - x^{2}}\,dx$$
