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Is there a reason of

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

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    $\begingroup$ It seems that $\cos(11x) + \sin(11(x+1)) \approx 0$ even if $x$ is not an integer, reaching a max and min of $\pm 0.0044$ (due to the fact that $11 \approx 3.5 \pi$). This is necessarily the best bound we can put on the value of $\cos(11n)+\sin(11(n+1))$, even if we specify $n \in 2\Bbb Z_+$. $\endgroup$ – Omnomnomnom Jan 9 '15 at 14:39
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    $\begingroup$ $11 \approx 3.5 \pi$, so $\sin(x) + \cos(x+11)$ is always close to zero. $\endgroup$ – Arthur Jan 9 '15 at 14:42
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    $\begingroup$ @Arthur which is similar to saying $\pi \approx \dfrac{22}{7}$ $\endgroup$ – Henry Jan 9 '15 at 14:44
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    $\begingroup$ It holds equally well for negative $n$, too. $\endgroup$ – Henning Makholm Jan 9 '15 at 14:45
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Looking at the unit circle, one can see that $\sin(x+\frac{3\pi}{2}) = -\cos(x)$. Using the fact that trig functions are $2\pi$ periodic, $\sin(x+\frac{7\pi}{2})$ also equals $-\cos(x)$. Now, $\sin(11(n+1)) = \sin(11n + 11)$, and $11 \approx \frac{7\pi}{2}$, so $\sin(11n+11)\approx -\cos(11n)$. This relation in fact holds for any $n$, not just for integers.

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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 $$ \begin{align} \left|\cos(11x)+\sin(11(x+1))\right| &\le\sqrt{(1-\cos(\delta))^2+\sin^2(\delta)}\\[6pt] &=2|\sin(\delta/2)|\\ &=2\left|\sin\left(\frac{22-7\pi}4\right)\right|\\ &=0.004425708823779\tag{3} \end{align} $$

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To make this slightly more explicit: let $f(n) = \cos(11n) + \sin(11(n+1))$. Then you can write

$$f(n) = \cos (11n) + \sin (11n+11) $$

and using the addition formula for sine

$$f(n) = \cos (11n) + \sin 11n \cos 11 + \cos 11n \sin 11. $$

Rearranging a bit this is

$$f(n) = (\cos 11n) (1 + \sin 11) + (\sin 11n) \cos 11. $$

Now since $7\pi/2 \approx 11$, we have $\sin 11 \approx \sin 7\pi/2 = -1$ and $\cos 11 \approx \cos 7\pi/2 = 0$.

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