Showing that $\frac{\sin(a_{n-1}) + 1}{2}$ is a Cauchy sequence In my homework set, I have the following question:

Show that    $$ a_n = \frac{\sin(a_{n-1}) + 1}{2}, \quad a_1=0 $$
satisfies the definition of Cauchy sequence.

As we went over the concept of Cauchy sequences a bit too quickly in class, I'm puzzled about how I should go about showing this. Conceptually, I understand what a Cauchy sequence is, and I know that in $\mathbb{R}$ it is the same as convergence, but I find it hard to apply to the above sequence.
I'd appreciate some hints about how to approach this problem.
Edit: I've been working on this with help from responses below so I thought I'd update my work. It's verbose but I thought that might show my thinking process better.
My work so far:
$$
\begin{align}
|a_{n+1} - a_n| &= \left | \frac{\sin(a_n) + 1}{2} -  \frac{\sin(a_{n-1}) + 1}{2} \right | \\
\\
&= \frac{1}{2} \left | \sin(a_n) + 1 -  (\sin(a_{n-1}) + 1)\right | \\
\\
&= \frac{1}{2} \left | \sin(a_n) -  \sin(a_{n-1}) \right | \\
\\
&= \frac{1}{2} \left | 2 \sin(\frac{a_n - a_{n-1}}{2}) \times \cos(\frac{a_n + a_{n-1}}{2}) \right | \\
\\
&\leq \frac{1}{2} \left | \sin(\frac{a_n - a_{n-1}}{2}) \times \cos(\frac{a_n + a_{n-1}}{2}) \right | \\
\\
&\leq \frac{1}{2} \left | \sin(\frac{a_n - a_{n-1}}{2}) \right | \\
\\
&\leq \frac{\left | a_n - a_{n-1} \right |}{2}
\end{align}
$$
This set of inequalities work because $|sin(x)| \leq |x|, \ \forall x \in \mathbb{R}$.
Hence, we have:
$$
\begin{align}
|a_{n+1} - a_n| &\leq \frac{1}{2} \left |a_n - a_{n-1} \right | \leq \frac{1}{2^2} \left | a_{n-1} - a_{n-2} \right | \leq \dots \leq \frac{1}{2^{n-1}} \left |a_{2} - a_{1} \right |
\end{align}
$$
So we learn that the distance between consecutive terms is becoming smaller, and that:
$$
\begin{align}
\left | a_{n+1} - a_n \right | &\leq \frac{1}{2^{n-1}} \times \left | \frac{1}{2} - 0 \right | = \frac{1}{2} \times \frac{1}{2^{n-1}} = \frac{1}{2^n}
\end{align}
$$
Now, for $n>m$ and $k=m$ we have:
$$
\begin{align}
\left | a_n - a_m \right | &= \sum_{k=m}^{n-1} \left | a_{k+1} - a_k \right | \\
\\
&= \sum_{k=m}^{n-1} \frac{1}{2^k} \\
\\
&= \left ( \frac{1}{2^m} + \frac{1}{2^{m+1}} + \dots + \frac{1}{2^{n-2}} + \frac{1}{2^{n-1}} \right ) \\
\\
&= \frac{1}{2^{m-1}} \left ( \frac{1}{2} + \frac{1}{2^2} + \dots + \frac{1}{2^{n-m-2}} + \frac{1}{2^{n-m-1}} \right ) \\
\\
&\leq \frac{1}{2^{m-1}}
\end{align}
$$
So, if we have $|a_n - a_{n-1}| < \epsilon$, then $\frac{1}{2^{m-1}} < \epsilon$. If we solve for $m$, we get that $m > \frac{\log(\frac{2}{\epsilon})}{\log(2)}$ and therefore $N > \frac{\log(\frac{2}{\epsilon})}{\log(2)}$.
So we have found $|a_n - a_m| < \epsilon$ for $n, m > N$. Hence, $(a_n)$ is Cauchy.
Resources used:


*

*Answers given below.

*This video was very helpful

*Understanding the definition of Cauchy sequence

*Proving that a sequence such that $|a_{n+1} - a_n| \le 2^{-n}$ is Cauchy

*Showing a recursive sequence is Cauchy

*How do I find the limit of the sequence $a_n=\frac{n\cos(n)}{n^2+1}$ and prove it is a Cauchy sequence?
 A: Hint. Note that the function $f:\mathbb{R}\to \mathbb{R}$,
$$f(x)=\frac{\sin(x)+1}{2}$$
is a contraction: for any $x,y\in \mathbb{R}$ there exists $t$ between $x$ and $y$ such that
$$|f(x)-f(y)|\leq \left|\frac{\cos(t)}{2}\right||x-y|\leq \frac{|x-y|}{2}\tag{1}$$
where we used the Mean value theorem. Hence, for $n\geq 1$, by using (1) $(n-1)$ times, we obtain
$$|a_{n+1}-a_n|\leq \frac{1}{2}|a_{n}-a_{n-1}|\leq \dots\leq \frac{1}{2^{n-1}}|a_{2}-a_{1}|.$$
Can you take it from here?
P.S. If the Mean value theorem is not allowed then we may note that by the sum-to-product identity, 
$$\frac{\sin(x)+1}{2}-\frac{\sin(y)+1}{2}=\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right).$$
Since $|\sin(t)|\leq |t|$ and $|\cos(t)|\leq 1$, it follows that (1) holds.
A: I think it is not necessary to define such a function defined by Robert Z.
$\displaystyle |a_{n+1}-a_n|\le \frac{1}{2}|a_n-a_{n-1}|$ for all $n=1,2,\cdots$
So , $\displaystyle |a_{n+1}-a_n|\le \frac{1}{2}.|a_n-a_{n-1}|\le \frac{1}{2^2}.|a_{n-1}-a_{n-2}|\le \cdots \le \frac{1}{2^{n-1}}.|a_2-a_1| \text{ for all } n=1,2,\cdots$ 
Now , for any positive integer $p$ , $$|a_{n+p}-a_n|\\\le |a_{n+p}-a_{n+p-1}|+\cdots +|a_{n+2}-a_{n+1}|+|a_{n+1}-a_n|\\\le \left( \frac{1}{2^{n+p-2}}+\cdots+\frac{1}{2^{n}}+\frac{1}{2^{n-1}}\right).|a_2-a_1|\\=\frac{1}{2^{n-2}}.[1-(1/2)^p].|a_2-a_1|\\ \to 0 \text { , as } n\to \infty \text{ for all } p=1,2,\cdots $$
So , $\{a_n\}$ is Cauchy.
