1.Cho tam giác ABC vuông tại A (AB <AC ). Đường trung tuyến AM, đường cao AH, BC = 2a.
CMR : a, [tex]sin2\alpha[/tex][tex]= 2sin\alpha cos\alpha[/tex]
b, [tex]1+cos^{2}\alpha =2cos^{2}\alpha[/tex]
c, [tex]1-cos^{2}\alpha =2sin^{2}\alpha[/tex]
Đề bài câu b, c là cos2alpha chứ??????
Kẻ AH _|_ BC. Đặt [tex]\angle C=\alpha[/tex]

Khi đó:
[tex]sin\alpha =\frac{AH}{AC}, cos\alpha =\frac{CH}{AC},\angle AMH=2\alpha[/tex] (góc ngoài tam giác AMC cân có AM = MC = a)
[tex]\Rightarrow cos2\alpha =\frac{HM}{AM}=\frac{HM}{a}, sin2\alpha =\frac{AH}{AM}=\frac{AH}{a}[/tex]
a) [tex]2sin\alpha cos\alpha =2.\frac{AH}{AC}.\frac{CH}{AC}=\frac{2AH.CH}{AC^2}=\frac{2AH.CH}{BC.CH}=\frac{AH}{a}=sin2\alpha (dpcm)[/tex]
b) [tex]2cos^2\alpha =2(\frac{CH}{AC})^2=\frac{2CH^2}{BC.CH}=\frac{CH}{a}\\ 1+cos2\alpha =1+\frac{HM}{a}=\frac{MC+HM}{a}=\frac{HC}{a}\\ \Rightarrow 2cos^2\alpha =1+cos2\alpha[/tex]
c) Ta có:
[tex]1+cos2\alpha =2cos^2\alpha \Rightarrow 1+cos2\alpha =2(1-sin^2\alpha )\Leftrightarrow 1+cos2\alpha =2-2sin^2\alpha \Leftrightarrow 2sin^2\alpha =1-cos2\alpha[/tex]