Sum to Infinity of Trigonometry to $\pi$ For 
$$y=\sum_{n=0}^a2\cdot2^n\cdot\tan\left(\frac{45}{2^n}\right)\cdot\sin\left(\frac{90}{2^n}\right)^2$$ 
I am currently working on a proof with a good friend of mine that involves adding more and more triangles to the sides of a regular polygon but keeping the longest diagonal constant until eventually, it becomes a circle.  And we ended up with this formula.
4-sided regular→ 8-sided regular→ 16-sided regular→ 32-sided regular→... →n-sided regular 
(When n tends to infinity, the area will be equal to that of a circle with the longest diagonal as diameter)
We have already tried Geometric Sequence Infinite Sum, but there does not seem to have a common ratio.
Moreover, we have used our calculator to input the numbers up to 128[(sin ...] and we get the value of 3.140... which is very close to π But we can't be completely sure that the infinity sum really equals to π.  
That is why we really need your knowledge of Maths to solve this.
Is there a way to prove that when $a$ tends to infinity, $y$ tends to $\pi$? 
Thanks in advance and Happy Holidays Everyone! :D

Hint:
$$2\tan A\cdot\sin^22A=2(4\sin^3A\cos A)=(3\sin A-\sin3A)2\cos A$$
$$=3\sin2A-(\sin4A+\sin2A)=2\sin2A-\sin4A$$
which clearly shows Telescopic form
