Maximum value of combined sinusoids

Question

How to find the maximum value of $2 \cdot \sin(3000\pi t)+3 \cdot \sin(4000\pi t)+7 \cdot \cos(6000\pi t)$?
I know can do this by the usual process like finding the first derivate with respect to $t$ but the process is way too tedious. Is there a neat and clean shortcut method which gives you the answer?

Answer 1

I don't think there is an easy solution or even an analytical solution.

Computing the derivative and rewriting the trigonometric functions in terms of $z=\exp(i1000\pi t)$, you get the ugly equation $$6\frac{z^3+z^{-3}}{2}+12\frac{z^4+z^{-4}}{2}-42\frac{z^6-z^{-6}}{2i}=0.$$

By looking at a plot of the derivative (in green), you see that it does have all $12$ roots per period.

Answer 2

It's not at all neat and clean. For convenience let $x = 1000 \pi t$, so you want the maximum of $f(x) = 2 \sin(3x) + 3 \sin(4x) + 7 \cos(6x)$. Now $f'(x) = 6 \cos(3x) + 12 \cos(4x) - 42 \sin(6x)$. We need to look at the zeros of that function, and pick out the one which maximizes $f$. Numerical methods appear to be necessary for this, but let's see how far we can get algebraically.

Expanding it out, with $s = \sin(x)$ and $c = \cos(x)$ we have $f'(x) = -1344 c^5 s+96 c^4+1344 c^3 s+24 c^3-96 c^2-252 c s-18 c+12$, where $c^2 + s^2 = 1$. The resultant of $f'(x)$ and $c^2 + s^2 - 1$ with respect to $s$ is $$36\, \left( 1+2\,c \right) ^{2} \left( 12544\,{c}^{10}-12544\,{c}^{9}- 28224\,{c}^{8}+31360\,{c}^{7}+18096\,{c}^{6}-25904\,{c}^{5}-696\,{c}^{ 4}+7116\,{c}^{3}-1723\,{c}^{2}-28\,c+4 \right) $$ which we need to be $0$. The easy solution $c = -1/2$ turns out not to give the maximum, so what we need is one of the roots of the irreducible $10$'th degree polynomial $$12544\,{c}^{10}-12544\,{c}^{9}-28224\,{c}^{8}+31360\,{c}^{7}+18096\,{c }^{6}-25904\,{c}^{5}-696\,{c}^{4}+7116\,{c}^{3}-1723\,{c}^{2}-28\,c+4 $$ Yes, numerical methods are required.

Answer 3

Considering $$y=2 \sin(3000\pi t)+3 \sin(4000\pi t)+7\cos(6000\pi t)$$ start changing variable $x=6000 \pi t$. This gives $$y=2 \sin \left(\frac{x}{2}\right)+3 \sin \left(\frac{2 x}{3}\right)+7 \cos (x)$$ What you can expect is that the maximum would occur "close" to $x=0$. So, use a Taylor expansion which gives $$y=7+3 x-\frac{7 x^2}{2}+O\left(x^3\right)$$ $$y'=3-7 x+O\left(x^2\right)$$ Using this approximation, the derivative cancels for $x=\frac 37$ (the second derivative test confirms that this is a maximum); at this point you could easily obtain the exact solution using Newton method using $x_0=\frac 37$ for the equation $$\cos \left(\frac{x}{2}\right)+2 \cos \left(\frac{2 x}{3}\right)-7 \sin (x)=0$$ I report below the successive iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.428571428571428571428571428571 \\ 1 & 0.426680172118602358382697690638 \\ 2 & 0.426680643919175965732057636382 \\ 3 & 0.426680643919205192593040906812 \\ 4 & 0.426680643919205192593040906925 \end{array} \right)$$

Now, go back to $t$ if required and compute $y$.