Hint: Use these two facts and a little bit of algebra:
$$\lim_{x \to 0} \frac{1-\cos(x)}{x} = 0$$
$$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$$
Note that $$\lim_{x \to 0} 10 \times \frac{1-\cos(2x)}{2x} = 0$$
The most straightforward and rigorous way to see why the first relation holds is to use the Taylor series for cosine. A non-rigorous way of seeing the first relation holds, if we accept the second relation, besides using the double-angle formula and similar trigonometric relations as pointed out by others, is this non-rigorous but more or less intuitive argument:
When $x$ is small, we know that:
$$\sin(x) \approx x $$
$$\cos(x)=\int -\sin(x) \approx -x^2/2+C$$
If $x=0$, then $\cos(x)=1$, hence $C=1$.
This shows that $1-\cos(x)=x^2/2$ when $x$ is small and this proves the first relation.
Consider the sequence $x_n$ defined by $x_0 = x$ and $x_{n+1} = \cos(x_n)$. Then the sequence in question is
$$a_n = n\prod_{n=1}^\infty x_n.$$
I claim that $a_n \to 0$. Here's a sketch of the proof:
- There is a unique point $x^* \in [0, 1)$ such that $\cos(x^*) = x^*$ (the fixed point of $\cos$).
- The sequence of fixed point iterates $x_n$ converge to $x^*$, regardless of the value of $x$.
- The sum $\sum a_n$ converges, using the ratio test.
- The sequence $a_n$ converges to $0$, using the divergence test.
The hard bit is 2, which I'll leave to last. To prove 1, note that the function $f(x) = x - \cos(x)$ is continuous, negative at $0$, and positive at $\pi/2$, so by the intermediate value theorem, there must be at least one point where $f(x) = 0$ in $[0, \pi/2]$.
Further, $f'(x) = 1 + \sin(x) \ge 0$, meaning that the function is non-decreasing. If $f$ had more than one root, then it'd be an interval of roots, which would correspond to an interval of roots in $f'$. This is clearly not the case, so there is a unique $x^*$ such that $f(x^*) = 0$, i.e. $\cos(x^*) = x^*$.
The point $x^*$ lies in the range of $\cos$, i.e. $[-1, 1]$, as well as $[0, \pi/2]$, so $x^* \in [0, 1]$. If $x^* = 1$, then $\cos(x^*) = 1$, hence $x^*$ would have to be an integer multiple of $2\pi$, which it is clearly not. Thus, $x^* \in [0, 1)$, as claimed.
To prove 3, assuming 2 is proven, consider
$$\left|\frac{a_{n+1}}{a_n}\right| = \frac{n+1}{n}|x_{n+1}| \to x^* < 1,$$
thus the series converges absolutely. Then, 4 follows immediately from this: the terms of a convergent series must tend to $0$.
Now, we tackle 2. First, recall the trigonometric identity:
$$\cos(x) - \cos(y) = -2\sin\left(\frac{x + y}{2}\right)\sin\left(\frac{x - y}{2}\right).$$
Now, suppose that $x, y \in [0, 1]$. Note that $\frac{x + y}{2} \in [0, 1]$ and $\sin$ is increasing and positive on $[0, 1] \subseteq [0, \pi/2]$, hence
$$\left|\sin\left(\frac{x + y}{2}\right)\right| = \sin\left(\frac{x + y}{2}\right) \le \sin(1).$$
Also, recall that $|\sin \theta| \le |\theta|$ for all $\theta$. Hence, assuming still $x, y \in [0, 1]$,
$$|\cos(x) - \cos(y)| = 2\left|\sin\left(\frac{x + y}{2}\right)\sin\left(\frac{x - y}{2}\right)\right| < 2 \cdot \sin(1) \cdot \left| \frac{x - y}{2}\right| = \sin(1)|x - y|.$$
Now, note that $x_n \in [-1, 1]$ for $n \ge 1$, and since $\cos$ is positive over $[-1, 1]$, we have $x_n \in [0, 1]$ for $n \ge 2$. So, for $n \ge 2$, we get
$$|x_{n+2} - x_{n+1}| = |\cos(x_{n+1}) - \cos(x_n)| \le \sin(1)|x_{n+1} - x_n|.$$
This implies the series
$$\sum_{n=2}^\infty (x_{n+1} - x_n)$$
is absolutely summable, as it passes the ratio test (limsup version):
$$\left|\frac{x_{n+2} - x_{n+1}}{x_{n+1} - x_n}\right| = \frac{|x_{n+2} - x_{n+1}|}{|x_{n+1} - x_n|} \le \sin(1) \frac{|x_{n+1} - x_n|}{|x_{n+1} - x_n|} = \sin(1) < 1.$$
Therefore, $\sum_{n=2}^\infty (x_{n+1} - x_n)$ converges. This is a telescoping series, whose partial sums take the form $x_n - x_2$. These partial sums converge, and hence so must $x_n$.
Now, because $x_n$ converges to some $L$, it follows from $\cos$ being continuous that
$$x_{n+1} = \cos(x_n) \implies L = \cos(L) \implies L = x^*,$$
completing step 2, and the full proof, as necessary.
( I think even desmos couldn't compute it further)
Best Answer
If $x$ is a rational multiple of $\pi$, then for some integer $N > 0$, $\sin(Nx) = 0$. This forces $\prod_{n=1}^r \sin(nx) = 0$ whenever $r \ge N$. In this case, the limit is $0$.
Otherwise, $x$ is not a rational multiple of $\pi$ and $|\cos x| < 1$. Notice
$$|\sin(nx)\sin(n+1)x| = \frac{|\cos x - \cos((2n+1)x)|}{2} \le \mu \stackrel{def}{=}\frac{1 + |\cos x|}{2}$$
By grouping the factors in the numerator in pairs, we have following bound for the weighted product at finite $r$.
$$r\left|\prod_{n=1}^r \sin(nx)\right| \le r\prod_{k=1}^{\lfloor r/2\rfloor} |\sin((2k-1)x)\sin(2kx)| \le r\mu^{\lfloor r/2\rfloor} $$ Since $\mu < 1$, the limit of the weighted product is again $0$.