Giải thích các bước giải:
Ta có :
$A=\sin x\cos x$
$\to 2A+1=1+2\sin x\cos x$
$\to 2A+1=\sin^2x+\cos^2x+2\sin x\cos x$
$\to 2A+1=(\sin x+\cos x)^2$
$\to 2A+1=m^2$
$\to 2A=m^2-1$
$\to A=\dfrac12(m^2-1)$
Lại có :
$B=|\sin^4x-\cos^4x|$
$\to B=|\sin^2x-\cos^2x|\cdot |\sin^2x+\cos^2x|$
$\to B=|(\sin x-\cos x)(\sin x+\cos x)|\cdot 1$
$\to B=|(\sin x-\cos x)(\sin x+\cos x)|$
$\to B=|(\sin x-\cos x)\cdot m|$
$\to B=|m|\cdot\sqrt{(\sin x-\cos x)^2}$
$\to B=|m|\cdot\sqrt{(\sin x+\cos x)^2-4\sin x\cdot \cos x}$
$\to B=|m|\cdot\sqrt{m^2-4\cdot\dfrac12(m^2-1)}$
$\to B=|m|\cdot |-m^2+2|$
Lại có :
$m=\sin x+\cos x$
$\to |m|=|\sin x+\cos x|$
$\to |m|=\sqrt{(\sin x+\cos x)^2}$
$\to |m|\le \sqrt{2(\sin^2x+\cos^2x)}$
$\to |m|\le \sqrt{2}$
