Chứng minh đẳng thức:
\(\dfrac{2sin2x+sin4x}{2\left(cosx+cos3x\right)}\)=\(tan2x.cosx\)
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TN
1
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 12 2020
\(\dfrac{1+cosx+cos2x+cos3x}{2cos^2x+cosx-1}=\dfrac{1+cosx+2cos^2x-1+4cos^3x-3cosx}{2cos^2x+cosx-1}\)
\(=\dfrac{4cos^3x+2cos^2x-2cosx}{2cos^2x+cosx-1}=\dfrac{2cosx\left(2cos^2x+cosx-1\right)}{2cos^2x+cosx-1}=2cosx\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 8 2021
\(cos^3xsinx-sin^3xcosx=sinx.cosx\left(cos^2x-sin^2x\right)=\dfrac{1}{2}sin2x.cos2x=\dfrac{1}{4}sin4x\)
\(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=1-\dfrac{1}{2}\left(2sinx.cosx\right)^2=1-\dfrac{1}{2}sin^22x\)
\(=1-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)=\dfrac{3}{4}+\dfrac{1}{4}cos4x=\dfrac{1}{4}\left(3+cos4x\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 5 2020
\(\frac{cos^3x-cos3x}{cosx}+\frac{sin^3x+sin3x}{sinx}=cos^2x-\frac{cos3x}{cosx}+sin^2x+\frac{sin3x}{sinx}\)
\(=1+\frac{sin3x.cosx-cos3x.sinx}{sinx.cosx}=1+\frac{sin\left(3x-x\right)}{\frac{1}{2}sin2x}=1+\frac{2sin2x}{sin2x}=3\)
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 6 2020
\(\frac{1+sin4x+cos4x}{1-sin4x+cos4x}=\frac{1+2sin2x.cos2x+2cos^22x-1}{1-2sin2x.cos2x+2cos^22x-1}\)
\(=\frac{2cos2x\left(sin2x+cos2x\right)}{2cos2x\left(cos2x-sin2x\right)}=\frac{sin2x+cos2x}{cos2x-sin2x}\)
\(=\frac{\sqrt{2}sin\left(2x+\frac{\pi}{4}\right)}{\sqrt{2}cos\left(2x+\frac{\pi}{4}\right)}=tan\left(2x+\frac{\pi}{4}\right)\)
\(\left(sin5x-cos5x\right)^2-\left(sin3x+cos3x\right)^2\)
\(=\left(\sqrt{2}sin\left(5x-\frac{\pi}{4}\right)\right)^2-\left(\sqrt{2}sin\left(3x+\frac{\pi}{4}\right)\right)^2\)
\(=2sin^2\left(5x-\frac{\pi}{4}\right)-2sin^2\left(3x+\frac{\pi}{4}\right)\)
\(=1-cos\left(10x-\frac{\pi}{2}\right)-1+cos\left(6x+\frac{\pi}{2}\right)\)
\(=-sin10x-sin6x=-2sin8x.cos2x\)
5 tháng 7 2021
1,\(VT=\dfrac{sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}{cos\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}+\dfrac{cos\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}{sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}\)\(=\dfrac{sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)^2+cos^2\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}{cos\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}\)
\(=\dfrac{1}{\dfrac{1}{2}.sin\left(\dfrac{\pi}{2}+x\right)}=\dfrac{2}{cosx}=VP\)
2,\(VT=\left(sin^4x-cos^4x\right)\left(sin^4x+cos^4x\right)=\left(sin^2x+cos^2x\right)\left(sin^2x-cos^2x\right)\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)
\(=\left(sin^2-cos^2x\right)\left(1-2sin^2x.cos^2x\right)\)\(=-cos2x\left(1-\dfrac{1}{2}sin^22x\right)\)\(=-\dfrac{cos2x\left(2-sin^22x\right)}{2}=-\dfrac{cos2x\left(1+cos^22x\right)}{2}\)
\(VP=-\left(\dfrac{7}{8}cos2x+\dfrac{1}{8}cos6x\right)=-\dfrac{7}{8}cos2x-\dfrac{1}{8}\left[4cos^32x-3cos2x\right]=-\dfrac{7}{8}.cos2x-\dfrac{1}{2}cos^32x+\dfrac{3}{8}cos2x\)
\(=-\dfrac{1}{2}cos2x-\dfrac{1}{2}cos^32x=\dfrac{-cos2x\left(1+cos^22x\right)}{2}\)
\(\Rightarrow VT=VP\)(đpcm)
3, \(VT=3-4\left(1-2sin^2x\right)+1-2sin^22x=8sin^2x-2sin^22x=8sin^2x-8.sin^2x.cos^2x=8sin^2x\left(1-cos^2x\right)=8sin^4x=VP\)
4,\(VP=\dfrac{1}{2}\left[sin\left(x+\dfrac{\pi}{2}\right)+sin\left(3x+\dfrac{\pi}{6}\right)\right]-\dfrac{1}{2}\left[cos\left(3x-\dfrac{\pi}{3}\right)+cos\left(x+\pi\right)\right]\)
\(=\dfrac{1}{2}\left(cosx+sin3x.\dfrac{\sqrt{3}}{2}+\dfrac{cos3x}{2}\right)-\dfrac{1}{2}\left(\dfrac{cos3x}{2}+sin3x.\dfrac{\sqrt{3}}{2}-cosx\right)\)
\(=\dfrac{1}{2}.2cosx=cosx=VP\)
5, \(VP=4cos\left(2x-\dfrac{\pi}{6}\right).\left(sinx.\dfrac{\sqrt{3}}{2}+\dfrac{cosx}{2}\right)^2\)\(=cos\left(2x-\dfrac{\pi}{6}\right).\left(sinx.\sqrt{3}+cosx\right)^2\)
\(VT=2.cos\left(2x-\dfrac{\pi}{6}\right)+2.sin\left(2x-\dfrac{\pi}{6}\right).cos\left(2x-\dfrac{\pi}{6}\right)=2cos\left(2x-\dfrac{\pi}{6}\right)\left[1+sin\left(2x-\dfrac{\pi}{6}\right)\right]\)
\(=2cos\left(2x-\dfrac{\pi}{6}\right)\left(1+\dfrac{sin2x.\sqrt{3}}{2}-\dfrac{cos2x}{2}\right)\)\(=2cos\left(2x-\dfrac{\pi}{6}\right)\left(sin^2x+cos^2x+sinx.cosx.\sqrt{3}-\dfrac{cos^2x-sin^2x}{2}\right)\)
\(=2cos\left(2x-\dfrac{\pi}{6}\right)\left(sin^2x.\dfrac{3}{2}+sinx.cosx.\sqrt{3}+\dfrac{cos^2x}{2}\right)\)\(=cos\left(2x-\dfrac{\pi}{6}\right)\left(sin^2x.3+2sinx.cosx.\sqrt{3}+cos^2x\right)\)
\(=cos\left(2x-\dfrac{\pi}{6}\right)\left(sinx.\sqrt{3}+cosx\right)^2\)
\(\Rightarrow VT=VP\) (dpcm)
RT
1
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 11 2018
\(sin4x+1-2sinx-sin2x-cos3x=0\)
\(\Leftrightarrow2cos3x.sinx-cos3x+1-2sinx=0\)
\(\Leftrightarrow cos3x\left(2sinx-1\right)-\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left(cos3x-1\right)\left(2sinx-1\right)=0\Rightarrow\left[{}\begin{matrix}cos3x=1\\sinx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(VT=\dfrac{2\cdot sin2x+2\cdot sin2x\cdot cos2x}{2\cdot\left(cosx+cos3x\right)}\)
\(=\dfrac{2\cdot sin2x\left(1+cos2x\right)}{2\cdot\left(cosx+cos3x\right)}\)
\(=\dfrac{sin2x\cdot\left(1+2cos^2x-1\right)}{cosx+cos3x}\)
\(=\dfrac{sin2x\cdot2\cdot cos^2x}{2\cdot cos\left(\dfrac{3x+x}{2}\right)\cdot cos\left(\dfrac{3x-x}{2}\right)}\)
\(=\dfrac{sin2x\cdot cos^2x}{cosx\cdot cos2x}=\dfrac{sin2x}{cos2x}\cdot cosx=tan2x\cdot cosx\)