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.
It is enough to study $f(x)=\left|\sin(x)\right|+\left|\sin(x+1)\right|$ over $[0,\pi]$ to be able to state that:
$$\forall x\in\mathbb{R},\qquad \left|\sin(x)\right|+\left|\sin(x+1)\right|\geq \sin(1). $$
Equality is attained at $x\in\pi\mathbb{Z}$ and at $x\in -1+\pi\mathbb{Z}$. 

We may also improve the upper bound in the following way:
$$\forall x\in\mathbb{R},\qquad \left|\sin(x)\right|+\left|\sin(x+1)\right|\leq 2\cos\left(\frac{1}{2}\right). $$
Equality is attained at $x=\frac{\pi-1}{2}+\pi\mathbb{Z}$.