Chứng minh các đẳng thức sau:
3. sin^4x-cos^4x=1-2cos^2x
4. 1-cosx/sinx=sinx/1+cosx
5.[sinx/1+cosx]+[1+cosx/sinx]=2/sinx
6.sinx+cosx-1/1-cosx=2cosx/sinx-cosx+1
$VT=\sin^4x-\cos^4x\\=(\sin^2x+\cos^2x)(\sin^2x-\cos^2x)\\=(\sin^2x+\cos^2x)(\sin^2x+\cos^2x-2\cos^2x)\\=1-2\cos^2x=VP(do\ \sin^2x+\cos^2x=1)$
Giả sử
$\dfrac{1-\cos x}{\sin x}=\dfrac{\sin x}{1+\cos x}\\\Leftrightarrow \dfrac{1-\cos^2x}{\sin x+\sin x\cos x}=\dfrac{\sin^2x}{\sin x+\sin x\cos x}\\\Leftrightarrow \dfrac{\sin^2x}{\sin x+\sin x\cos x}=\dfrac{\sin^2x}{\sin x+\sin x\cos x}(luôn\ đúng)$
Vậy ...
$VT=\dfrac{\sin x}{1+\cos x}+\dfrac{1+\cos x}{\sin x}\\=\dfrac{\sin^2x}{\sin x+\sin x\cos x}+\dfrac{1+2\cos x+\cos^2x}{\sin x+\sin x\cos x}\\=\dfrac{2+2\cos x}{\sin x+\sin x\cos x}(do\ \sin^2x+\cos^2x=1)\\=\dfrac{2(1+\cos x)}{\sin x(1+\cos x)}\\=\dfrac{2}{\sin x}=VP$
$\dfrac{\sin x+\cos x-1}{1-\cos x}\\=\dfrac{(\sin x+\cos x-1)(\sin x-\cos x+1)}{(1-\cos x)(\sin x-\cos x+1)}\\=\dfrac{\sin^2x-(1-\cos x)^2}{(1-\cos x)(\sin x-\cos x+1)}\\=\dfrac{\sin^2x-1 +2\cos -\cos^2x}{(1-\cos x)(\sin x-\cos x+1)}\\=\dfrac{2\cos x-2\cos^2x}{(1-\cos x)(\sin x-\cos x+1)}(do\ \sin^2x+\cos^2x=1)\\=\dfrac{(2\cos x)(1-\cos x)}{(1-\cos x)(\sin x-\cos x+1)}\\=\dfrac{2\cos x}{\sin x-\cos x+1}=VP$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
