25~26高三上·湖南湘东联盟·第14题

$1$

【题目】设$\alpha, \beta \in \left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$，若$\alpha^3 + \cos\left(\alpha + \frac{3\pi}{2}\right) – 2t = 0$、$4\beta^3 + \sin\beta\cos\beta + t = 0$，则$\cos(\alpha + 2\beta) =$___________．

【答案】$1$

【解析】$\alpha^3 + \cos\left(\alpha + \frac{3\pi}{2}\right) – 2t = 0$，
化简得$\alpha^3 + \sin\alpha = 2t$，①
$4\beta^3 + \sin\beta\cos\beta + t = 0$，
$\therefore 8\beta^3 + 2\sin\beta\cos\beta + 2t = 0$，
得$(2\beta)^3 + \sin2\beta = -2t$，②
$\because \alpha, \beta \in \left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$，
$\therefore 2\beta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$，
令$f(x) = x^3 + \sin x, x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$，
$\therefore f'(x) = 3x^2 + \cos x$，
$\because 3x^2 \geq 0, \cos x \geq 0$，
$\therefore f'(x) \geq 0, f(x)$在定义域上单调递增，
由①，②可得$f(\alpha) = 2t, f(2\beta) = -2t$，
又$\because f(x)$是奇函数，
$\therefore \alpha = -2\beta$，即$\alpha + 2\beta = 0$，
$\therefore \cos(\alpha + 2\beta) = \cos0 = 1$.
故答案为：$1$.