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 $AC\perp BC, SA\perp BC\to BC\perp SAC$

$\to \widehat{SBC,ABC}=\widehat{SCA}$

$\to \cos\widehat{SCA}=\dfrac{CA}{SC}=\sqrt{\dfrac23}$

$\to \widehat{SCA}=\arccos\left(\sqrt{\dfrac23}\right)$

$\to \alpha=\arccos\left(\sqrt{\dfrac23}\right)$

$\to \sin2\alpha=\dfrac{2\sqrt{2}}{3}$