Cho tam giác ABC vuông tại A, AB < AC, co...

- Hướng dẫn giải

Góc 2α = $\widehat{AMH}$

a, Ta có: $\sin 2\alpha =\frac{AH}{AM}=\frac{2AH}{AM}$ = $2\frac{AB.AC}{B{C}^2}=2\sin \alpha .\cos \alpha$

b,  1 + cos2α = $1+\frac{HM}{AM}=\frac{HC}{AM}=\frac{2HC}{BC}$ = $2.\frac{A{C}^2}{B{C}^2}=2{\cos }^2\alpha$

c, 1 – cos2α = $1-\frac{HM}{AM}=\frac{HB}{AM}=\frac{2HB}{BC}$ = $2.\frac{A{B}^2}{B{C}^2}=2{\sin }^2\alpha$