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For a constant, N, what value of x will maximize the cosine (or any trig) function?

\begin{equation} 1 = \cos{(Nx)} \end{equation}

I am looking for the exact form, not the approximation because, \begin{equation} \frac{\arccos{(1)}}{N} = x = 0 \end{equation}

For example, WolframAlpha.com states that if N = 19.013, then, \begin{equation} x = \frac{2000 \pi n}{19013} , n \text{ } \varepsilon \text{ } \text{set of integers} \end{equation} How was that solution calculated?

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    How do you maximize an equation? You can _solve_ an equation. The solution comes from examining the set of angles with cosine 1, which you can figure out by looking at the unit circle.2010-11-30
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    I am solving for when cosine is maximized which is one2010-11-30
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    Try first the case N=1, then N=2, then see if you can decide the general process. (Polya: if you cannot answer a question, there is a simpler question you also cannot answer. Answer that one first)2010-11-30
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    It is true the maximum value of cosine is 1, do you know for what x cosine attains its maximum?2010-11-30
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    @WWright, `cosine(pi) = 1` but I need to solve this in terms of `N`2010-11-30
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    Maybe there's a value of $x$ that works for all $N$?2010-11-30
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    I have just a quick note on your equation stating, \begin{equation} x = \frac{2000 \pi n}{19013} , n \text{ } \varepsilon \text{ } \text{set of integers} \end{equation} You should typically $\in$ (\in) rather than $\varepsilon$. Also, to make the $\mathbb{Z}$ symbol, you type \mathbb{Z}. The reason I mention this is because I used to typeset the symbols as you have above and then realized the proper $\LaTeX$ code to get what I want.2010-11-30
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    @Elpezmuerto $\cos\left(\pi\right)=-1$ but $\cos\left(0\right)=1$ and cosine is a periodic function, do you see where to take it from here?2010-11-30
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    Use binary search.2010-11-30
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    -1...please explain2010-11-30

3 Answers 3

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Cosine takes it's maximum when the argument is $2k\pi$, where $k$ is any integer. Therefore

$$\begin{aligned}2k\pi = Nx \\\\ x=\frac{2k\pi}{N}\end{aligned}$$

Thus for $N=19.013$ $$x= \frac{2k\pi}{19.013} = \frac{2000k\pi}{19013}$$.

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We know that

\begin{equation} 1 = \cos{(2\pi)} \end{equation}

Therefore, for a given N to "maximize" cosine:
\begin{aligned} 2\pi = Nx \ \end{aligned} \begin{aligned} x = \frac{2\pi}{N} \end{aligned}

Using the above equation: \begin{aligned} x &= \frac{2\pi}{19.013} \ \end{aligned} \begin{aligned} x &\approx 0.33046 \ \end{aligned}

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    The WolframAlpha answer is more general, but only because it accounts for the fact that cos(t) = 1 for t every even multiple of pi (including zero).2010-11-30
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$x=0$ is always a solution, since $\cos(N\times0)=\cos0=1$ .

If you want the set of all solutions, use the fact that

$$\cos\theta=1 \Leftrightarrow \theta=2k\pi \text{ with } k\in\mathbb{Z}$$

Replacing $\theta$ by $Nx$ above gives you the set of solutions :

$$ 1=\cos{Nx}\Leftrightarrow x\in \left\{\frac{2k\pi}N\right\}_{k\in\mathbb{Z}}$$

This is the solution given by Wolfram Alpha for $N=19.013=\frac{19013}{1000}$.