Question: Consider:

$$\lim\limits\_{n\to +\infty}\sqrt[n]{|\sin(n+1)|+|\sin(n)|}=1$$

I figured out that
$$\lim\limits\_{n\to +\infty}\sqrt[n]{|\sin(n+1)|+|\sin(n)|}\leqslant\lim\_{n\to +\infty}\sqrt[n]{2}=1$$

What about the other side inequality?

Answer: 
We show that $|\sin(n+1)|+|\sin(n)|$ is never tiny, by finding a crude lower bound.

Suppose that $|\sin(n)|\lt 0.3$. By the addition formula, we have $\sin(n+1)=\sin n \cos 1+\cos n\sin 1$. Thus
$$|\sin(n+1)|\ge |\cos n\sin 1|-|\sin n\cos 1|.\tag{1}$$
We have $|\cos n|\gt \sqrt{1-(0.3)^2}\gt 0.8$ and since $\sin 1\gt 0.8$ we have $|\cos n\sin 1|\gt 0.64$. But $|\sin n\cos 1|\lt 0.3$, so from (1) we conclude that $|\sin(n+1)|\gt 0.3$.

We have shown that one at least of $|\sin(n+1)|$ and $|\sin(n)|$ is $\ge 0.3$, so the sum is greater than $0.3$. 

Finally, $\sqrt[n]{0.3}$ has limit $1$ as $n\to\infty$, and the squeeze is on.
You have $2>|\sin(n+1)|+|\sin(n)|>0$ is continuous and periodic, so there is an $a$ witch satisfies $a>0$ and $2>|\sin(n+1)|+|\sin(n)|>a$, now you can take the n-th root of a, bacause $\lim\_{n\to +\infty}\sqrt[n]{a}=1$