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I would like to know if there is a simple approach to find the range of functions in the form: $$\sin x\sin2x$$ $$\cos x\cos3x$$ $$\sin 2x\cos 4x$$

For example, finding the range of a function in the form: $$a\cos\theta + b\sin\theta$$ is simple (the minimum value is $-\sqrt{a^2 + b^2}$ while the maximum value is $\sqrt{a^2 + b^2}$.

Gummy bears
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    You could take the derivatives of them to find the locations of the minima and maxima and plug those values in. – turkeyhundt Jan 23 '15 at 05:46
  • Are the $x$ multipliers rationally related? – copper.hat Jan 23 '15 at 05:57
  • @turkeyhundt Well of course that would be the typical approach. I was hoping for something simpler? That would require some time. Isn't there a more direct approach? – Gummy bears Jan 23 '15 at 05:57
  • @copper.hat What do you mean by 'rationally related'? As in they are both rational numbers? – Gummy bears Jan 23 '15 at 05:58
  • All the above are of the form $f(ax)g(bx)$ where $f,g \in {\sin, \cos}$. I am asking if ${a \over b}$ is rational. – copper.hat Jan 23 '15 at 05:59
  • Yes $a/b$ is rational. And of course they are of that form, but differentiating will require some time nonetheless. Isn't there a direct formula? Even in $a\cos\theta + b\sin\theta$ we can apply differentiation. – Gummy bears Jan 23 '15 at 06:01
  • The range of $x \mapsto \sin x \sin(2x)$ is $\pm {4 \over 3 \sqrt{3}}$, but it takes a little work. – copper.hat Jan 23 '15 at 06:37
  • What if it was cosine instead of sine. And also, what if the coefficients of $x$ inside were different? @copper.hat – Gummy bears Jan 23 '15 at 06:47
  • I was just pointing out that there isn't a simple approach (if you are considering differentiation not to be simple). – copper.hat Jan 23 '15 at 07:01
  • @copper.hat So there is no other option but to differentiate? Also, looking at the graph, I don't think that the method of differentiating would work. There are two maximas, at different heights. Strange graph. – Gummy bears Jan 23 '15 at 07:02
  • If the $a,b$ are rationally related (as in the comment above) then the functions are periodic and smooth, so finding the zeroes of the derivative will certainly 'work'. – copper.hat Jan 23 '15 at 07:04
  • @copper.hat I was using this website (https://www.desmos.com/calculator) to graph the functions. Try graphing $2\cos x\sin 2x$. It has two peaks. – Gummy bears Jan 23 '15 at 07:08
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    Obviously the larger of the two maximas is the global maxima. Likewise for any minima. Since sine and cosine are continuous, their product is continuous, so you don't have to worry about any larger or smaller points. – A. Thomas Yerger Jan 23 '15 at 07:09
  • @AlfredYerger That means that I will get two values when I use the differentiating method? – Gummy bears Jan 23 '15 at 07:15
  • In general, you will get many possible values which you then have to evaluate and check. – copper.hat Jan 23 '15 at 07:16
  • I haven't done any computation, but it seems reasonable that there should be more than one peak. You just have to take the values, shove them into the function and compare the outputs. The larger one is your maxima, since these functions are continuous. Same idea for your minima. Note that this method of differentiation does not distinguish between maxima and minima - both have derivative 0. A sketch will aid you. – A. Thomas Yerger Jan 23 '15 at 07:17
  • @AlfredYerger Of course, but sketching such a graph by hand is quite difficult. Anyways thanks! – Gummy bears Jan 23 '15 at 07:20
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    Well it's not too bad, because you'll know where the derivative is positive or negative, and the shape of the graph, plus you'll be evaluating certain points to give your graph a pretty well defined shape. This standard process of curve sketching should be a part of every calculus course. If you're unsure what to do, just google "curve sketching calculus." – A. Thomas Yerger Jan 23 '15 at 16:11

3 Answers3

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It's more interesting when it is not the obvious upper bound $1$. I will take the example $\sin 3x \cos 5 x$. It does not reach the value $1$, so we have some work to do. Let's find an implicit equation for the curve $C \colon \{(\sin(3t), \cos (5t))\ \mid \ t\in [0, 2 \pi]\}$ ( http://en.wikipedia.org/wiki/Lissajous_curve). Skipping some details, it is the curve with equation $$-1 + 25 x^2 - 200 x^4 + 560 x^6 - 640 x^8 + 256 x^{10} + 9 y^2 - 24 y^4 + 16 y^6=0$$

It has to do with the Chebyshev polynomials. In fact, a point $(x,y)$ in $\mathbb{R}^2$ is of the form $(\sin (3t), \cos(5t))$ if and only if $1 = P(x) + Q(y)$ where $P(\sin( \alpha))= \sin^2 (5 \alpha)$ and $Q(\cos(\beta)) = \cos^2(3 \beta)$. (the only if is clear, since $\sin^2(15 t) + \cos^2(15 t) = 1$). So it is not that hard to get the implicit form for the curve $C$.

So now we need to find

$$\max x y \ \text{ where }\ -1 + 25 x^2 - 200 x^4 + 560 x^6 - 640 x^8 + 256 x^{10} + 9 y^2 - 24 y^4 + 16 y^6=0$$

We omit the calculations using Lagrange multipliers. It turns out that the maximum $M$ is the largest root of the equation $$1073741824\, t^8-1644167168\, t^6+656998400\, t^4-52537500 \,t^2+84375=0$$ $M= 0.96410...$

Well, at least setting the Lagrange multiplier problem in general is not that hard. Solving it is a different thing.

The Lissajous curve $C$

It is not clear to me whether a general easy formula for this maximum exists for general $m$, $n$ for $\max \sin(m t) \cos (n t )$. Maybe a general method, not a general formula that is easy to apply.

Alternatively one writes

$$\sin 3t \cos 5 t = \frac{1}{2}( \sin 8t - \sin 2t)$$

Reduce to an equivalent problem: maximize $ \sin 4u - \sin u$. Even this one is not straightforward. Certainly the derivative is easy to calculate but the maximal value is again the solution of an equation of degree $8$. Perhaps the advantage is that one can find the solution from the graphs of $\sin u$, $\sin 4u$. enter image description here

$\bf{Added:}$. If one looks for the maximum value of $\sin mt \cos nt$, it's enough to consider the case $m,n$ relatively prime. Then, even it the maximal value is not $1$, it will get closer to one with the increase of $\max (m,n)$. It is intuitive since the Lissajous curves tend to fill up the square. It would be interesting to investigate how close to $1$ one gets as $\max(m,n) \to \infty$. It appears that number theory, more precisely - rational approximation, appears.

orangeskid
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  • Sorry to say but all this is much much above my understanding capabilities. Might be helpful for others though. – Gummy bears Jan 23 '15 at 06:57
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    @Gummy bears: It helps if you have some mathematical software to plot curves in parametric and implicit form. No worries, at some point this will seem much easier. It is true that the standard approach with finding the zeroes of the derivative is not that simple in its calculations either. There may not be a neat formula for the maximum, just methods. – orangeskid Jan 23 '15 at 07:07
  • Oh... well thanks for the help. It was kind of scary just looking at all this. – Gummy bears Jan 23 '15 at 07:11
  • @Gummy bears: I propose you try to do $\max \sin t \cos 3t$ with different methods, maybe with the above method ( find the Lissajou curve $(\sin t ,\cos 3t)$, plot it, see where the curve is tangent to some hyperbola $x y = $ const, this example may look easier after that. – orangeskid Jan 23 '15 at 07:20
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    @Gummy bears: The calculations were done with Mathematica, computer algebra systems can be useful at times. – orangeskid Jan 23 '15 at 07:24
  • Hmmmm..... but I would like to do it without the aid of computer or calculator. – Gummy bears Jan 23 '15 at 07:55
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    @Gummy bears: You could use http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Product-to-sum_and_sum-to-product_identities, at least it becomes more intuitive – orangeskid Jan 23 '15 at 08:21
  • @Gummybears In particular, $\sin3t\cos5t=\frac12(\sin8t-\sin2t)$ is just an application of the product=to-sum rule $\sin u\cos v=\frac12[\sin(u+v)+\sin(u-v)]$ where $u=3t$ and $v=5t$. – David K Jan 23 '15 at 13:21
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i don't know if this what you are looking for.

take for example the fitst function $y = f(x) = \sin x \sin 2x.$ you have product of two periodic functions of periods $\pi$ and $2\pi.$ so the product is $\pi$-periodic. you can think of $y = \sin x \sin 2x = A\sin 2x $ where the amplitude $A = \sin x$ not constant but slowly varying compared to the faster varying $\sin 2x$ in fact the graph of $f(x)$ is between the envelopes $y = \pm \sin x$ you will see the local maximum of $f$ a little to the right of $\pi/4,$ local max of $\sin 2x$ and the local min a little to the left of $3\pi/4,$ local min of $\sin 2x.$

all this will be much clearer if you can sketch the graphs of $y = f(x), y = \pm \sin x.$

abel
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  • Yes it kind of becomes clear upon graphing it. However, I was actually looking for the value of the maximum of $f(x)$ – Gummy bears Jan 24 '15 at 06:58
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    @Gummybears, it is unlikely that you can find the exact maximum value without the use calculus. even with calculus, you can only find approximate values because you will be solving a trig equation. – abel Jan 24 '15 at 08:33
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We have always $ a^2\leq a^2+b^2\ \Rightarrow\ |a|\leq\sqrt{a^2+b^2} \Rightarrow \frac{|a|}{\sqrt{a^2+b^2}}\leq 1\Rightarrow-1\leq\frac{a}{\sqrt{a^2+b^2}}\leq1$ so there exists a $\alpha $ that $ \cos\alpha=\frac{a}{\sqrt{a^2+b^2}}$. Beside We know that $\sin\alpha=\pm\sqrt{1-\cos^2\alpha}=\pm\sqrt{1-(\frac{a}{\sqrt{a^2+b^2}})^2}=\pm\frac{b}{\sqrt{a^2+b^2}}$. Now let be $y=a cos \theta+b\sin \theta $ then $\frac{y}{\sqrt{a^2+b^2}}=\frac{a}{\sqrt{a^2+b^2}}\sin \theta+\frac{b}{\sqrt{a^2+b^2}}\cos \theta=\cos\alpha\sin \theta\pm\sin\alpha\cos \theta=\sin (\theta\pm\alpha)$. Because of $-1\leq\sin (\theta\pm\alpha)\leq 1$ then $-1\leq\frac{y}{\sqrt{a^2+b^2}}\leq 1$. Hence $-\sqrt{a^2+b^2}\leq a\cos \theta+b\sin \theta\leq \sqrt{a^2+b^2}$.

bigli
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