0
$\begingroup$

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?

  • 0
    -1...please explain2010-11-30

3 Answers 3

1

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

0

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}

  • 0
    The WolframAlpha answer is more general, bu$t$ only because it accounts for the fact that cos(t) = 1 for t every even multiple of pi (including zero).2010-11-30
0

$x=0$ 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 $N=19.013=\frac{19013}{1000}$.