2sinx-sin2x/2 sinx+sin2x=sin²x/(1+cosx)² Cm hai vế = nhau 17/10/2021 Bởi Valentina 2sinx-sin2x/2 sinx+sin2x=sin²x/(1+cosx)² Cm hai vế = nhau
Giải thích các bước giải: $VT=\dfrac{2\sin x-\sin2x}{2\sin+\sin2x }\\=\dfrac{2\sin x-2\sin x\cos x}{2\sin+2\sin x\cos x }\\=\dfrac{2\sin x(1-\cos x)}{2\sin x(1+\cos x) }\\=\dfrac{1-\cos x}{1+\cos x}\\VP=\dfrac{\sin^2x}{(1+\cos x)^2}\\=\dfrac{1-\cos^2x}{(1+\cos x)^2}\\=\dfrac{(1-\cos x)(1+\cos x)}{(1+\cos x)^2}\\=\dfrac{1-\cos x}{1+\cos x}=VT\Rightarrow ĐPCM$ Bình luận
$VT=\dfrac{2\sin x -\sin 2x}{2\sin x+\sin 2x}$ $=\dfrac{2\sin x -2\sin x.\cos x}{2\sin x+ 2\sin x.\cos x}$ $=\dfrac{2\sin x(1-\cos x)}{2\sin x(1+\cos x)}$ $=\dfrac{1-\cos x}{1+\cos x}$ $=\dfrac{(1-\cos x)(1+\cos x)}{(1+\cos x)^2}$ $=\dfrac{1-\cos^2x}{(1+\cos x)^2}$ $=\dfrac{\sin^2x}{(1+\cos x)^2}$ $= VP$ Bình luận
Giải thích các bước giải:
$VT=\dfrac{2\sin x-\sin2x}{2\sin+\sin2x }\\
=\dfrac{2\sin x-2\sin x\cos x}{2\sin+2\sin x\cos x }\\
=\dfrac{2\sin x(1-\cos x)}{2\sin x(1+\cos x) }\\
=\dfrac{1-\cos x}{1+\cos x}\\
VP=\dfrac{\sin^2x}{(1+\cos x)^2}\\
=\dfrac{1-\cos^2x}{(1+\cos x)^2}\\
=\dfrac{(1-\cos x)(1+\cos x)}{(1+\cos x)^2}\\
=\dfrac{1-\cos x}{1+\cos x}=VT\Rightarrow ĐPCM$
$VT=\dfrac{2\sin x -\sin 2x}{2\sin x+\sin 2x}$
$=\dfrac{2\sin x -2\sin x.\cos x}{2\sin x+ 2\sin x.\cos x}$
$=\dfrac{2\sin x(1-\cos x)}{2\sin x(1+\cos x)}$
$=\dfrac{1-\cos x}{1+\cos x}$
$=\dfrac{(1-\cos x)(1+\cos x)}{(1+\cos x)^2}$
$=\dfrac{1-\cos^2x}{(1+\cos x)^2}$
$=\dfrac{\sin^2x}{(1+\cos x)^2}$
$= VP$