Is this function a non-zero function? Assume that $(a_i,b_i)^T \in \mathbb{R}^2\setminus \{ (0,0) \}$ for $i=1, \dots, n$ and let $s_1, \dots, s_n \in \mathbb{R}\setminus \{0 \}$ be distinct constants. Define the function
$$
f(t) = \sum_{i=1}^{n} a_i\sin(s_it) - b_i \cos(s_it).
$$
Is this a nonzero function? It looks very much nonzero to me but somehow I cannot prove this. One just has to find a value $t$ such that $f(t) \neq 0$
 A: This is a case where proving stronger results is easier than giving a formula for a $t$ at which this function is zero.
The most classical approach to showing that this function is not zero is to, essentially, look at its Fourier transform. In elementary terms, all we need to do is choose some $i$ and look at, over a suitable large range, the following integral:
$$\int_{-R}^Rf(t)\cdot \left(a_i\sin(s_it) - b_i\cos(s_it)\right)$$
If you bring the summation for $f$ outside the integral, you will see that the terms you are interested in are those of the form
$$\int_{-R}^R\left(a_j\sin(s_jt) - b_j\cos(s_jt)\right)\cdot \left(a_i\sin(s_it) - b_i\cos(s_it)\right)$$
When $j=i$, this term grows linearly with $R$ since it is just the square of a non-zero periodic function. When $j\neq i$, the integrand is a sum of terms* of the form $\alpha \sin((s_j\pm s_i) t + \omega)$ where the coefficients can be determined via trigonometric identities (or by writing everything in terms of $e^{i\theta}$) and its integral is bounded with respect to $R$. Thus, you find that this integral is a sum of something that grows linearly with $R$ plus some bounded terms - hence for large enough $R$, it is, in particular, not zero. This implies that $f$ is not zero.
Another approach is to look at its derivatives (at $0$, if we need concreteness). The $n^{th}$ derivative will be, for large enough $n$, dominated by the highest frequency term - in particular, the derivatives of $a_i\sin(s_it)-b_i\cos(s_it)$ grow on the order of $s_i^n$ (admitting that they may be zero for half the coefficients if $a_i=0$ or $b_i=0$) - and this rate of growth outpaces any lower frequencies - a nice corollary of this is that, since this actually applies everywhere, this implies that the function never has every derivative vanish, which implies, in turn, that the roots are actually isolated - so there's at most countably many zeros (and, indeed, since, due to your assumption that no $s_i=0$, the integral of the function over any large interval is bounded, there are indeed infinitely many roots)

*To be more explicit about this step, you can either look at the following four trigonometric identities:
$$\cos(a)\cos(b) = \frac{1}2\left(\cos(a+b) + \cos(a-b)\right)$$
$$\sin(a)\cos(b) = \frac{1}2\left(\sin(a+b) + \sin(a-b)\right)$$
$$\cos(a)\sin(b) = \frac{1}2\left(\sin(a+b) - \sin(a-b)\right)$$
$$\sin(a)\sin(b) = \frac{1}2\left(\cos(a+b) - \cos(a-b)\right)$$
You can of course derive all of these from the identity $e^{i\theta} = \cos(\theta) + i\sin(\theta)$ and if you were doing a lot of calculations like this, you would definitely use this form instead. In any case, note that if we have $a=s_it$ and $b=s_jt$, then every term we get in the product $$(a_j\sin(s_jt) - b_j\cos(s_jt))\cdot (a_i\sin(s_i t) - b_i \cos(s_i t))$$
is a constant times either the sine or cosine of $(s_i\pm s_j)t$. So long as $s_i\neq \pm s_j$, this implies that this product expands to a sum of sines and cosines with non-zero angular frequency - but the integral of both $\sin(\alpha t)$ and $\cos(\alpha t)$ are bounded whenever $\alpha$ is not zero, so the integral of these terms must also be bounded.
