Đáp án: $ \sin2\alpha=\dfrac{2\sqrt{2}}{3}$
Giải thích các bước giải:
Gọi $AC=x,x>0$
Vì $\Delta ABC$ vuông cân tại $C\to CA=CB, CB\perp AB$
Ta có $SA\perp ABC\to SA\perp AC$
$\to SA=\sqrt{SC^2-AC^2}=\sqrt{a^2-x^2}$
$\to V_{SABC}=\dfrac13\cdot SA\cdot \dfrac12\cdot AC\cdot BC=\dfrac16\cdot SA\cdot AC^2$
$\to V_{SABC}=\dfrac16\cdot \sqrt{a^2-x^2}\cdot x^2$
Đặt $f\left(x\right)=\sqrt{a^2-x^2}\cdot x^2$
$\to f'\left(x\right)=\dfrac{-2x}{2\sqrt{a^2-x^2}}\cdot x^2+\sqrt{a^2-x^2}\cdot 2x$
$\to f'\left(x\right)=0$
$\to \dfrac{-2x}{2\sqrt{a^2-x^2}}\cdot x^2+\sqrt{a^2-x^2}\cdot 2x=0$
$\to \dfrac{-1}{2\sqrt{a^2-x^2}}\cdot x^2+\sqrt{a^2-x^2}=0$
$\to x^2-2\left(a^2-x^2\right)=0$
$\to 3x^2=2a^2$
$\to x=\sqrt{\dfrac23}a$
$\to $Hàm số đạt cực đại tại $x=\sqrt{\dfrac23}a$
$\to f\left(x\right)\le \dfrac{2a^3}{3\sqrt{3}}$
$\to V_{SABC}\le \dfrac16\cdot \dfrac{2a^3}{3\sqrt{3}}$
$\to V_{SABC}\le \dfrac{a^3\sqrt{3}}{27}$
Dấu = xảy ra khi $x=\sqrt{\dfrac23}a$
Ta có $AC\perp BC, SA\perp BC\to BC\perp SAC$
$\to \widehat{SBC,ABC}=\widehat{SCA}$
$\to \cos\widehat{SCA}=\dfrac{CA}{SC}=\sqrt{\dfrac23}$
$\to \widehat{SCA}=\arccos\left(\sqrt{\dfrac23}\right)$
$\to \alpha=\arccos\left(\sqrt{\dfrac23}\right)$
$\to \sin2\alpha=\dfrac{2\sqrt{2}}{3}$
