Problem - 2254
Let $x, y \in [-\frac{\pi}{4}, \frac{\pi}{4}], a \in \mathbb{Z}^+$, and
$$
\left\{
\begin{array}{rl}
x^3 + \sin x - 2a &= 0 \\
4y^3 +\frac{1}{2}\sin 2y +a &=0
\end{array}
\right.
$$
Compute the value of $\cos(x+2y)$
The given equations are equivalent to
$$
\left\{
\begin{array}{rl}
x^3 + \sin x &= 2a \\
(-2y)^3 +\sin (-2y) &=2a
\end{array}
\right.
$$
Since $x, y \in [-\frac{\pi}{4}, \frac{\pi}{4}]$, we have $x, -2y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
Let $f(t) = t^3 + \sin t$, we have $f(x) = f(-2y)$. Because $f(t)$ is monotonically increasing in $[-\frac{\pi}{2}, \frac{\pi}{2}]$, we must have $x=-2y$. Hence $\cos(x+2y)=\cos 0 = 1$.