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\(P=sin^22x-\left[2sin\dfrac{x}{2}cos\dfrac{x}{2}\left(cos^4\dfrac{x}{2}-sin^4\dfrac{x}{2}\right)\right]^2\)
\(=sin^22x-\left[sinx\left(cos^2\dfrac{x}{2}-sin^2\dfrac{x}{2}\right)\left(cos^2\dfrac{x}{2}+sin^2\dfrac{x}{2}\right)\right]^2\)
\(=sin^22x-\left[sinx.cosx.1\right]^2\)
\(=sin^22x-\left[\dfrac{1}{2}sin2x\right]^2\)
\(=\dfrac{3}{4}sin^22x=\dfrac{3}{4}\left(1-cos^22x\right)=\dfrac{3}{4}\left(1-\dfrac{1}{4}\right)=\dfrac{9}{16}\)
\(sinx+cosx=m\Leftrightarrow\left(sinx+cosx\right)^2=m^2\)
\(\Leftrightarrow1+2sinx.cosx=m^2\Rightarrow sinx.cosx=\dfrac{m^2-1}{2}\)
\(A=sin^2x+cos^2x=1\)
\(B=sin^3x+cos^3x=\left(sinx+cosx\right)^3-3sinx.cosx\left(sinx+cosx\right)\)
\(=m^3-\dfrac{3m\left(m^2-1\right)}{2}=\dfrac{2m^3-3m^3+3m}{2}=\dfrac{3m-m^3}{2}\)
\(C=\left(sin^2+cos^2x\right)^2-2\left(sinx.cosx\right)^2=1-2\left(\dfrac{m^2-1}{2}\right)^2\)
\(D=\left(sin^2x\right)^3+\left(cos^2x\right)^3=\left(sin^2x+cos^2x\right)^3-3\left(sin^2x+cos^2x\right)\left(sinx.cosx\right)^2\)
\(=1-3\left(\dfrac{m^2-1}{2}\right)^2\)
\(=\dfrac{tan\left(\dfrac{pi}{2}+x\right)\cdot sin\left(-x\right)\cdot cos\left(x-pi\right)}{cos\left(\dfrac{pi}{2}-x\right)\cdot sin\left(x+pi\right)}\)
\(=\dfrac{-cotx\cdot sin\left(-x\right)\cdot\left(-cosx\right)}{sinx\cdot-sinx}\)
\(=\dfrac{cotx\cdot sinx\left(-1\right)\cdot cosx}{-sinx\cdot sinx}=\dfrac{\dfrac{cosx}{sinx}\cdot cosx}{sinx}=\dfrac{cos^2x}{sin^2x}=cot^2x\)
a: \(sinx+cosx=\sqrt{2}\)
=>\(\left(sinx+cosx\right)^2=2\)
=>\(1+2\cdot sinx\cdot cosx=2\)
=>\(2\cdot sinx\cdot cosx=1\)
=>\(sinx\cdot cosx=\dfrac{1}{2}\)
b: \(\left(sinx-cosx\right)^2=\left(sinx+cosx\right)^2-4\cdot sinx\cdot cosx\)
\(=2-4\cdot\dfrac{1}{2}=2-2=0\)
=>\(sinx-cosx=0\)
c: \(sinx-cosx=0\)
\(sinx+cosx=\sqrt{2}\)
Do đó: \(sinx=cosx=\dfrac{\sqrt{2}}{2}\)
\(\left(sinx+cosx\right)^2=\frac{25}{16}\Rightarrow1+2sinx.cosx=\frac{25}{16}\)
\(\Rightarrow sinx.cosx=\frac{9}{32}\)
\(\left(sinx-cosx\right)^2=\left(sinx+cosx\right)^2-4sinx.cosx=\frac{25}{16}-4.\frac{9}{32}=\frac{7}{16}\)
\(\Rightarrow sinx-cosx=\pm\frac{\sqrt{7}}{4}\)
số GP a đẹp cực><Nguyễn Việt Lâm