Giải thích các bước giải:

Ta có:

$f(x)=\sin^{2025}(x+2024\pi)\cos(x-\dfrac{2023\pi}2)$

$\to f(x)=\sin^{2025}(x)\cos(x-1012\pi+\dfrac12\pi)$

$\to f(x)=\sin^{2025}(x)\cos(x+\dfrac12\pi)$

$\to f(x)=\sin^{2025}(x)\sin(\dfrac{\pi}2-(x+\dfrac12\pi))$

$\to f(x)=\sin^{2025}(x)\sin(-x)$

$\to f(x)=-\sin^{2025}(x)\sin(x)$

Ta có:

$f(-x)=-\sin^{2025}(-x)\sin(-x)$

$\to f(-x)=-(-\sin^{2025}(x))\cdot (-\sin(x))$

$\to f(-x)=-\sin^{2025}(-x)\sin(-x)$

$\to f(-x)=f(x)$

$\to $Hàm số là hàm chẵn