# Evaluating $\sin\frac{\pi}{2}\sin\frac{\pi}{2^2}\sin\frac{\pi}{2^3}\cdots\sin\frac{\pi}{2^{11}}\cos \frac{\pi}{2^{12}}$

Evaluate :
$$\sin\frac{\pi}{2} \times \sin \frac{\pi}{2^2} \times \sin \frac{\pi}{2^3} \cdots \times \sin\frac{\pi}{2^{11}} \times \cos \frac{\pi}{2^{12}}$$
I tried to solve it by using double angle formula, by replacing $$\sin\displaystyle\frac{\pi}2$$ with $$2 \sin\displaystyle\frac{\pi}4 \cos\displaystyle\frac{\pi}4$$ and similarly replacing $$\sin\displaystyle\frac{\pi}{2^2}$$ with double angle formula. But I'm not getting anywhere from this, can you please suggest me how can I approach to this problem, especially that cosine term at last is making everything so weird.
The thing which is most important to ask is : although I know every theory which applies over here but then also why I'm unable to solve this question? What is the reason for my failure?

• Let me suggest that to improve the readability of your question, you set your formulas in mathjax. I did part of it for you. – Lee Mosher Jan 31 at 15:01
• @Jan What about the $\cos{{\pi\over12}}$ at the end? – saulspatz Jan 31 at 15:04
• Also, the reason for failing at something is usually that you haven't yet found the path to success. You could give up (sometimes a realistic choice), or you could keep trying. – Lee Mosher Jan 31 at 15:08
• One way is to convert the product into a sum. Hopefully, a summable series (i.e., expressible in closed form) would result. – Allawonder Jan 31 at 15:09
• – Cosmas Zachos Jan 31 at 15:24

Whatever you would do, the result would be a very small number very close to $$0$$ $$\sin\frac{\pi}{2} \times \sin \frac{\pi}{2^2} \times \sin \frac{\pi}{2^3}\times \cdots \times \sin\frac{\pi}{2^{11}} \times \cos \frac{\pi}{2^{12}}$$
We know the value of the sine for some of the angles. The first are $$\{1 , \frac {\sqrt 2} 2,\frac{\sqrt{2-\sqrt{2}}}{2},\frac{\sqrt{2-\sqrt{2+\sqrt{2}}}}{2},\frac{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}}{2},\frac{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}} }{2} \}$$ and the pattern continue for ever.
For the next one, corresponding to $$\sin \frac{\pi}{2^8}$$ the exact value is $$\frac{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}}}}{2}\approx0.012271538$$ while $$\frac{\pi}{2^8} \approx 0.012271846$$ that is to say that you can approximate all the sines by the argument. Concerning the cosine, the argument is so small that you can assume it is almost equal to $$1$$.
If you take into account the fact that $$\sin(x), an upper bound for the product of sines would be $$\frac{ \pi^9 \,\sqrt 2}{2^{64}}\approx 2.285 \times 10^{-15}$$ while it should be $$\approx 2.208 \times 10^{-15}$$.
Concerning $$\cos \frac{\pi}{2^{12}}\approx 0.99999970586$$