Áp dụng định lý Pytago, ta được:
$BC^2 = AB^2 + AC^2$
$\Rightarrow AB = \sqrt{BC^2 - AC^2} = \sqrt{a^2 - b^2}$
a) Ta có:
$sin\alpha = \dfrac{AB}{BC} = \dfrac{\sqrt{a^2 - b^2}}{a}$
$cos\alpha = \dfrac{AC}{BC} = \dfrac{b}{a}$
Do $M$ là trung điểm cạnh huyền $BC$
nên $MA= MB = MC = \dfrac{BC}{2} = \dfrac{a}{2}$
$\Rightarrow \widehat{MAC} = \widehat{MCA} = \alpha$
$\Rightarrow \widehat{AMB} = \widehat{MAC} + \widehat{MCA} = 2\alpha$
$\Rightarrow sin2\alpha = sin\widehat{AMB} = \dfrac{AH}{AM} = \dfrac{h_a}{\dfrac{a}{2}} = \dfrac{2h_a}{a}$
b) Ta có: $2sin\alpha.cos\alpha$
$=2.\dfrac{AH}{AC}.\dfrac{AC}{BC}$
$=\dfrac{2.AH}{BC}$
$=\dfrac{2AH}{2AM}$
$=\dfrac{AH}{AM}$
$=sin\widehat{AMB}$
$=sin2\alpha$
