Maximum value of combined sinusoids 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? 
 A: 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.

A: 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.  
