hard trigonometric equation.

[man111]

Let $\displaystyle f(x) = cos\left(a_{1}+x\right)+\frac{1}{2}cos\left(a_{2}+x\right)+\frac{1}{2^2}cos\left(a_{3}+x\right)+.............+\frac{1}{2^{n-1}}cos\left(a_{n}+x\right)$. Where $a_{1},a_{2},...........................a_{n}\in\mathbb{R}$.

If $f(x_{1}) = \f(x_{2}) = 0$. Then $\left|x_{2}-x_{1}\right| = $

[ИСН]

Look here, man. Expand them cosines, so the whole thing turns out as $a\cdot\cos x+b\cdot\sin x$, where $a$ and $b$ are merely two big ugly piles of junk. Then join them together again, like $\sqrt{a^2+b^2}\cdot\cos(x+\phi)$. Now, no matter what that $\phi$ may be, we know for sure that our peaks and zeros are precisely $\pi\over2$ apart. Well, $+\pi n$ or whatever.

[ewert]

ИСН в сообщении #427774 писал(а):

where $a$ and $b$ are merely two big ugly piles of junk

Пустячок, конечно, однако не хватает всё же доказательства того, что $a^2+b^2\neq0$.