TITLE: Is there a reason of $\cos(11x)+\sin(11(x+1))\approx 0$
QUESTION [9 upvotes]: Is  there a reason of 

$$\cos(11x)+\sin(11(x+1))\approx 0$$

REPLY [6 votes]: Note that
$$
11=\frac72\pi+\delta\tag{1}
$$
where $\delta=0.0044257124357$.
Therefore,
$$
\begin{align}
&\cos(11x)+\sin(11(x+1))\\[6pt]
&=\cos(11x)+\sin\left(11x+\frac72\pi+\delta\right)\\
&=\cos(11x)+\sin\left(11x+\frac72\pi\right)\cos(\delta)+\cos\left(11x+\frac72\pi\right)\sin(\delta)\\[6pt]
&=\cos(11x)-\cos(11x)\cos(\delta)+\sin(11x)\sin(\delta)\\[12pt]
&=\cos(11x)(1-\cos(\delta))+\sin(11x)\sin(\delta)\tag{2}
\end{align}
$$
Since the absolute value of a dot product is no more than the product of the lengths of the vectors, we get
$$
\begin{align}
\left|\cos(11x)+\sin(11(x+1))\right|
&\le\sqrt{(1-\cos(\delta))^2+\sin^2(\delta)}\\[6pt]
&=2|\sin(\delta/2)|\\
&=2\left|\sin\left(\frac{22-7\pi}4\right)\right|\\
&=0.004425708823779\tag{3}
\end{align}
$$