# Demonstrating an equivalent formula for $\sin(x)/x$ using power series

Would appreciate some ideas for the following:

"Prove that $$\frac{\sin{x}}{x}=\prod_{n=1}^{\infty}\cos{\frac{x}{2^n}}$$ using power series."

I'm aware this identity can be shown using trig identities and a telescoping product. Also, you can get another proof using the infinite product expressions $$\sin{x} = x\prod_{k=1}^{\infty} (1-\frac{x^2}{k^2 \pi^2})$$ and $$\cos{x}=\prod_{k=1, \ k \ \text{odd}}^\infty (1-\frac{4x^2}{k^2 \pi^2})$$.

However, since the question explicitly mentions power series, I was wondering if there is a proof that directly uses power series? I've tried calculating some derivatives and coefficients but they seem to get pretty nasty.

• use Taylor series for sin at zero? – Dole Dec 23 '18 at 6:26
• yeah, but the RHS is giving me trouble – Merk Zockerborg Dec 23 '18 at 6:28
• Is there a nice power series for $\log(\cos x)$? – mathworker21 Dec 23 '18 at 6:44

Using the fact that $$\sin(2x)=2\sin(x)\cos(x)$$ we have $$\cos(x)=\frac{\sin(2x)}{2\sin(x)}$$.
Define $$p_n(x):=\prod\limits_{j=1}^n \cos(\frac{x}{2^j})$$ and note that $$\cos(\frac{x}{2^j})=\dfrac{\sin(\frac{x}{2^{j-1}})}{2\sin(\frac{x}{2^j})}$$ for each $$j$$.
Now, $$p_n(x)=\frac{1}{2^n}\frac{\sin(x)}{\sin(\frac{x}{2^n})}=\frac{\sin(x)}{x}\dfrac{\frac{x}{2^n}}{\sin(\frac{x}{2^n})}$$ and $$p_n(x)\to_n \frac{\sin(x)}{x}$$
$$\cos z= e^{iz}(1+e^{-2iz})/2$$ so $$ln \cos z=\cdots$$