Estimating the sine of the angles between vectors

Question

Let $\{s_n\}$ be a sequence of unit vectors in $\mathbb{R}^2$ converging to some $s\in\mathbb{R}^2$. I want to show that $$\sin\measuredangle(s_n,s)\le \sum_{m=n}^\infty\sin\measuredangle(s_m,s_{m+1})$$ for all $n$, where all angles are taken to lie in $[0,\pi]$. This estimate makes intuitive sense from drawing a picture, but I am struggling to find a formal argument. Any suggestions would be appreciated.

Answer 1

Let $\theta_m = \measuredangle(s_{m}, s_{m+1})$. With an additional assumption that the series $\sum_{m=n}^{\infty}\theta_m$ converges (its sum is then equal to $\measuredangle(s_{n}, s)$ modulo $2\pi$) the proof simplifies a lot.

You need to show that: $$ \sin\left(\sum_{m=n}^{\infty}\theta_m\right) \leq \sum_{m=n}^{\infty} \sin(\theta_m). $$

First, let's prove the finite version of this inequality by induction with $$ \sin(\theta_n + \theta_{n+1}) = \sin(\theta_n)\cos(\theta_{n+1}) + \sin(\theta_{n+1})\cos(\theta_n) \leq \sin(\theta_n) + \sin(\theta_{n+1}). $$ as a basis. Then, having $$ \sin\left(\sum_{m=n}^{N}\theta_m\right) \leq \sum_{m=n}^{N} \sin(\theta_m). $$ proven for all $N > n$ take a limit with $N \to \infty$.

Answer 2

It looks to me like the main idea is that the derivative of sin is greatest near 0 (i.e. cosine is a decreasing function near 0). Let's try to solve this by following that intuition. It is not the most elegant solution but is a start.

$$\sin\measuredangle(a,b) = \int\limits_0^{|a-b|}\cos(\theta)d\theta$$ Then we want to get $$\int\limits_0^{|s_n-s|}\cos(\theta)d\theta \le \sum_{m=n}^\infty \int\limits_0^{|s_m-s_{m+1}|}\cos(\theta)d\theta$$

Then we break up the first integral to match the form of the second, so now we want to get $$ \int\limits_0^{|s_n-s|}\cos(\theta)d\theta =\sum_{m=n}^\infty \int\limits_{s_m}^{s_{m+1}}\cos(\theta)d\theta \le \sum_{m=n}^\infty \int\limits_0^{|s_m-s_{m+1}|}\cos(\theta)d\theta $$

For any term we can say, when angles are small, $$ \int\limits_{s_m}^{|s_{m+1}|}\cos(\theta)d\theta \le \int\limits_0^{|s_m-s_{m+1}|}\cos(\theta)d\theta $$

Which plugged in above finishes the 'proof'. This formal outline of our intuition works when the angles are small. It can be extended to cover everything, but you can probably find a clean proof if you look at sin(x) as a concave function: https://en.wikipedia.org/wiki/Concave_function

Answer 3

For the fixed $n$ and $m\ge n$, let $\theta_m \in (-\pi, \pi]$ be the signed angle from the $s_n$ to $s_m$,

$$\begin{align*} \theta_m &= \pm\measuredangle(s_n, s_m)\\ \measuredangle(s_n, s_m) &= |\theta_m| \end{align*}$$

Then incrementally, for either the positive or negative case,

$$\begin{align*} \measuredangle(s_k, s_{k+1}) &\equiv \pm(\theta_{k+1} - \theta_k) \pmod {2\pi}\\ \theta_{k + 1} &\equiv \theta_k \pm \measuredangle(s_k, s_{k+1}) \\ \sin \theta_{k+1}&= \sin\left[\theta_k \pm \measuredangle(s_k, s_{k+1})\right]\\ &= \sin \theta_k \cos \measuredangle(s_k, s_{k+1}) \pm \cos \theta_k \sin \measuredangle(s_k, s_{k+1})\\ \end{align*}$$

Taking absolute value on both sides,

$$\begin{align*} \left|\sin\theta_{k+1}\right| &= \left|\sin \theta_k \cos \measuredangle(s_k, s_{k+1}) \pm \cos \theta_k \sin \measuredangle(s_k, s_{k+1})\right|\\ &\le \left|\sin \theta_k \cos \measuredangle(s_k, s_{k+1})\right| + \left|\cos \theta_k \sin \measuredangle(s_k, s_{k+1})\right|\\ &= \left|\sin \theta_k\right|\left| \cos \measuredangle(s_k, s_{k+1})\right| + \left|\cos \theta_k\right|\left| \sin \measuredangle(s_k, s_{k+1})\right|\\ &\le \left|\sin \theta_k\right|+\left| \sin \measuredangle(s_k, s_{k+1})\right|\\ \sin\measuredangle(s_n, s_{k+1}) &\le \sin\measuredangle(s_n, s_k) + \sin \measuredangle(s_k, s_{k+1}) \end{align*}$$

By induction,

$$\sin\measuredangle(s_n, s_{k+1}) \le \sum_{m=n}^k \sin\measuredangle(s_m, s_{m+1})$$

Lastly, take $k\to \infty$ and $s = \lim_{k\to\infty} s_{k+1}$.

For convergence, the right hand side partial sums are non-negative and monotonic increasing. Regardless of whether the infinite sum diverges to infinity, it is still greater than the left hand side sine value.