Cho sinx + cosx = m Tính A = $sin^{3}$ + $cos^{3}$ B = $sin^{4}$ + $cos^{4}$ theo m 17/09/2021 Bởi Iris Cho sinx + cosx = m Tính A = $sin^{3}$ + $cos^{3}$ B = $sin^{4}$ + $cos^{4}$ theo m
$\quad \sin x + \cos x = m$ $\Rightarrow (\sin x +\cos x)^2 = m^2$ $\Rightarrow 1 + 2\sin x\cos x = m^2$ $\Rightarrow \sin x\cos x =\dfrac{m^2 -1}{2}$ Ta được: $+)\quad A = \sin^3x +\cos^3x$ $\to A = (\sin x +\cos x)^3 – 3\sin x\cos x(\sin x + \cos x)$ $\to A = m^3 – 3\cdot \dfrac{m^2 – 1}{2}\cdot m$ $\to A = \dfrac{3m – m^3}{2}$ $+)\quad B = \sin^4x +\cos^4x$ $\to B = (\sin^2x + \cos^2x)^2 – 2\sin^2x\cos^2x$ $\to B = 1 – 2\cdot \left(\dfrac{m^2 -1}{2}\right)^2$ $\to B = \dfrac{3 – m^2}{2}$ Bình luận
$\sin x+\cos x=m\to \sin^2x+\cos^2x+2\sin x\cos x=m^2\to \sin x\cos x=\dfrac{m^2-1}{2}$ $A=\sin^3x+\cos^3x=(\sin x+\cos x)(\sin^2x+\cos^2x-\sin x\cos x)$ $=(\sin x+\cos x)(1-\sin x\cos x)$ $=m.\Big(1-\dfrac{m^2-1}{2}\Big)$ $=m.\dfrac{2-m^2+1}{2}$ $=\dfrac{3m-m^3}{2}$ $B=\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x$ $=1-2(\sin x\cos x)^2$ $=1-2.\dfrac{(m^2-1)^2}{4}=1-\dfrac{(m^2-1)^2}{2}$ Bình luận
$\quad \sin x + \cos x = m$
$\Rightarrow (\sin x +\cos x)^2 = m^2$
$\Rightarrow 1 + 2\sin x\cos x = m^2$
$\Rightarrow \sin x\cos x =\dfrac{m^2 -1}{2}$
Ta được:
$+)\quad A = \sin^3x +\cos^3x$
$\to A = (\sin x +\cos x)^3 – 3\sin x\cos x(\sin x + \cos x)$
$\to A = m^3 – 3\cdot \dfrac{m^2 – 1}{2}\cdot m$
$\to A = \dfrac{3m – m^3}{2}$
$+)\quad B = \sin^4x +\cos^4x$
$\to B = (\sin^2x + \cos^2x)^2 – 2\sin^2x\cos^2x$
$\to B = 1 – 2\cdot \left(\dfrac{m^2 -1}{2}\right)^2$
$\to B = \dfrac{3 – m^2}{2}$
$\sin x+\cos x=m\to \sin^2x+\cos^2x+2\sin x\cos x=m^2\to \sin x\cos x=\dfrac{m^2-1}{2}$
$A=\sin^3x+\cos^3x=(\sin x+\cos x)(\sin^2x+\cos^2x-\sin x\cos x)$
$=(\sin x+\cos x)(1-\sin x\cos x)$
$=m.\Big(1-\dfrac{m^2-1}{2}\Big)$
$=m.\dfrac{2-m^2+1}{2}$
$=\dfrac{3m-m^3}{2}$
$B=\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x$
$=1-2(\sin x\cos x)^2$
$=1-2.\dfrac{(m^2-1)^2}{4}=1-\dfrac{(m^2-1)^2}{2}$