Weekly Challenge 5: Trigger

7063 /www/nrich/html/content/id/7063/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"> <ul id="buttonBar"> <li> <a href="http://nrich.maths.org/7071&part=">Try this next</a> </li> <li> <a href="http://plus.maths.org/content/frugal-nature-euler-and-calculus-variations">Read all about it</a> </li> <li> <a href="https://nrich.maths.org/discus/messages/27/27.html?1286783308">Ask NRICH</a> </li> <li> <a href="http://nrich.maths.org/6481&part=">Warm-up problem</a> </li> <li> <a href="http://nrich.maths.org/7032&part=">Last week's solution</a> </li> </ul> <div> Create a sketch of the function $y = \cos(\tan(x))$ for $-\pi \le x \leq \pi$, being sure to locate key turning points accurately. You might wish to attempt this without resorting to IT. </div> <div class="framework"> Did you know ... ? Whilst the properties of elementary functions might seem obvious, combinations of these can quickly lead to interesting behaviour which requires more thought to understand properly. Functions such as the one considered in the problem are part of the reason that calculus and functions are considered in more careful detail at university.</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> We are looking to sketch $y=\cos(\tan(x))$. What do we know about this? Certainly we will have $-1\leq y\leq 1$ since $\cos(x)$ is always between $-1$ and $1$, regardless of its arguament. Next, we know that $\cos(\theta)$ is zero for the values $\theta = (2n+1)\frac{\pi}{2}$ and has turning points inbetween these values at $m\pi$ Next, we know that $\tan{x}$ has asymptotes at $x = \pm \frac{\pi}{2}, \pm\frac{3\pi}{2}$. Only two of these lie in the range $-\pi \leq x \leq \pi$. This allows us to see that $\tan x$ takes every real number value exactly twice on the range $-\pi < x \leq \pi$ and is undefined at two points on this range. Thus, $y=\cos(\tan x)$ will have an infinite number of turning points on the specified range, and these occur precisely at the values $$ 0, \pm\tan^{-1}(\pi), \pm \tan^{-1} (2\pi), \pm \tan^{-1}(3\pi) \dots $$ Numerically, these values are $$ 0, \pm 1.262627, \pm 1.412965, \pm 1.465089, \pm 1.491386, \pm 1.50720, \pm 1.517794, \dots $$ Note that these values tend to the numbers $\pm \frac{\pi}{2}$, which are the two values for which our function is undefined Similarly, zeros will occur at the values $$ \pm\tan^{-1}\left(\frac{\pi}{2}\right),\tan^{-1}\left(\frac{3\pi}{2}\right), \tan^{-1}\left(\frac{5\pi}{2}\right)\dots $$ Putting all this together we obtain a bounded function which oscillates infinitely often in any interval containing the point $\frac{\pi}{2}$ or the point -$\frac{\pi}{2}$ The final part is to determine which points are maxima and which are minima. This is easily done by evaluating $\cos(\tan(0)) = +1$ and $\cos(\tan(\pm\pi))=1$. The key points can easily be indicated with paper and pencil, but here is the output from graphmatica so that you can check your answers. <mdo:image height="415" width="521" src="output.png" alt=""></mdo:image> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> To sketch any graph look for: zeros; turning points; asymptotes; positive values and negative values; the evaluation of the function at key points, such as $0$, $\pm 1$ or the boundaries of the range or domain of the function (i.e. the possible input values and max/min output values). Also, ask yourself: Does the graph vary smoothly from point to point? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> We are looking to sketch $y=\cos(\tan(x))$. What do we know about this? Certainly we will have $-1\leq y\leq 1$ since $\cos(x)$ is always between $-1$ and $1$, regardless of its arguament. Next, we know that $\cos(\theta)$ is zero for the values $\theta = (2n+1)\frac{\pi}{2}$ and has turning points inbetween these values at $m\pi$ Next, we know that $\tan{x}$ has asymptotes at $x = \pm \frac{\pi}{2}, \pm\frac{3\pi}{2}$. Only two of these lie in the range $-\pi \leq x \leq \pi$. This allows us to see that $\tan x$ takes every real number value exactly twice on the range $-\pi < x \leq \pi$ and is undefined at two points on this range. Thus, $y=\cos(\tan x)$ will have an infinite number of turning points on the specified range, and these occur precisely at the values $$ 0, \pm\tan^{-1}(\pi), \pm \tan^{-1} (2\pi), \pm \tan^{-1}(3\pi) \dots $$ Numerically, these values are $$ 0, \pm 1.262627, \pm 1.412965, \pm 1.465089, \pm 1.491386, \pm 1.50720, \pm 1.517794, \dots $$ Note that these values tend to the numbers $\pm \frac{\pi}{2}$, which are the two values for which our function is undefined Similarly, zeros will occur at the values $$ \pm\tan^{-1}\left(\frac{\pi}{2}\right),\tan^{-1}\left(\frac{3\pi}{2}\right), \tan^{-1}\left(\frac{5\pi}{2}\right)\dots $$ Putting all this together we obtain a bounded function which oscillates infinitely often in any interval containing the point $\frac{\pi}{2}$ or the point -$\frac{\pi}{2}$ The final part is to determine which points are maxima and which are minima. This is easily done by evaluating $\cos(\tan(0)) = +1$ and $\cos(\tan(\pm\pi))=1$. The key points can easily be indicated with paper and pencil, but here is the output from graphmatica so that you can check your answers. <mdo:image height="415" width="521" src="output.png" alt=""></mdo:image> </mdoxml> 2 3 0 0 0 0 1 Weekly Challenge 5: Trigger A weekly challenge: these are shorter problems aimed at Post-16 students or enthusiastic younger students. Collections Weekly Challenge Trigonometry Trigonometry generally Admin Stage 5 - Core Mapping Stage 5 Core Mapping Document Trigonometry AS Secondary Mapping Document DisplayCabinet