a) Ta có
$\sin x + \cos x = m$
Bình phương 2 vế ta có
$\sin^2x + \cos^2x + 2\sin x \cos x = m^2$
$<-> 1 + 2\sin x \cos x = m^2$
$<-> \sin x \cos x = \dfrac{m^2-1}{2}$
Vậy $\sin x \cos x = \dfrac{m^2-1}{2}$
Ta có
$|\sin^4x - \cos^4x| = |(\sin^2x - \cos^2x)(\sin^2x + \cos^2x)|$
$= |\sin^2x - \cos^2x|$
$= |(\sin x - \cos x)(\sin x + \cos x)|$
$= |m(\sin x - \cos x)|$
Ta có
$(\sin x - \cos x)^2 = (\sin x + \cos x)^2 - 4\sin x \cos x$
$= m^2 - 4. \dfrac{m^2-1}{2}$
$= 2 - m^2$
Suy ra $\sin x - \cos x = \sqrt{2 - m^2}$
Vậy
$|\sin^4x - \cos^4x| = |m \sqrt{2-m^2}|$
b) Ta có
$m = \sin x + \cos x$
$= \sqrt{2} \left( \dfrac{1}{\sqrt{2}} \sin x + \dfrac{1}{\sqrt{2}} \cos x \right)$
$= \sqrt{2} (\sin x \cos 45 + \cos x \sin 45)$
$= \sqrt{2} \sin(x + 45)$
Ta có
$|\sin (x + 45)| \leq 1$ với mọi $x$
$<-> \sqrt{2} |\sin(x+45)| \leq \sqrt{2}$ với mọi $x$
$<-> |m| \leq \sqrt{2}$ với mọi $x$.
