Let $a_1, a_2, \dots, a_n$ (real numbers) be such that
$$a_1 - a_2/3 + \dots + (-1)^{n - 1}a_n/(2n - 1) = 0.$$
Prove that
$$f(x):= a_1\cos(x) + a_2\cos(3x) + \dots + a_n\cos([2n - 1]x) = 0,$$
for some $x \in (0, \pi/2)$.
Base case (n = 1): This implies that $a_1 = 0$ (by hypothesis), so that $f(x) = a_1\cos(x) = 0$.
Now assume that the claim holds for $n = k \in \mathbb{N}$. Must show that
$$f(x) = a_1\cos(x) + a_2\cos(3x) + \dots + a_k\cos([2k - 1]x) + a_{k + 1}\cos([2k + 1]x) = 0.$$
By the inductive hypothesis, we have that
$$f(x) = a_1\cos(x) + a_2\cos(3x) + \dots + a_k\cos([2k - 1]x) = 0.$$
Hence,
$$a_{k + 1}\cos([2k + 1]x) = 0.$$
If $a_{k + 1} = 0$, we are done. Suppose not. Then
$$\cos([2k + 1]x) = 0 \implies x = \frac{\pi/2}{2k + 1}.$$
Thus, we have shown that
$$f(x):= a_1\cos(x) + a_2\cos(3x) + \dots + a_n\cos([2n - 1]x) = 0,$$
for some $x \in (0, \pi/2)$.
Are there any analysis tools that one could use to approach this problem differently?
The $\frac 1{2k-1}$ smell like integration. Note that $\int a_k\cos([2k-1]x)\,\mathrm dx=-\frac1{2k-1}\sin([2k-1]x)$. So consider $$F(s):=\sum_k \frac{a_k}{2k-1}\sin([2k-1]x)$$ and apply Rolle's theorem