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3) \(\frac{sin2x-sinx}{1-cosx+cos2x}=tanx\)
4) \(\left(\frac{sinx+cotx}{1+sinx.tanx}\right)^{2014}=\frac{sin^{2014}x+cot^{2014}x}{1+sin^{2014}x.tan^{2014}x}\)
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HN
29 tháng 4 2020
\(a,\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4sin^2x.cos^2x}=-1\)
\(VT=\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4.sin^2x.cos^2x}=\left(\frac{1}{tan2x}\right)^2-\frac{1}{sin^22x}=\left(\frac{cos2x}{sin2x}\right)^2-\frac{1}{sin^22x}=\frac{cos^22x-1}{sin^22x}=\frac{-sin^22x}{sin^22x}=-1=VP\)
b, \(VT=\frac{cos^2x-sin^2x}{sin^4x+cos^4x-sin^2x}=\frac{cos2x}{\left(sin^2x+cos^2x\right)^2-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{1-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{cos^2x-2.sin^2x.cos^2x}\)
=\(\frac{cos2x}{cos^2x.\left(1-2.sin^2x\right)}=\frac{cos2x}{cos^2x.cos2x}=\frac{1}{cos^2x}=1+tan^2x=VP\)
d, \(VT=\left(\frac{cosx}{1+sinx}+tanx\right).\left(\frac{sinx}{1+cosx}+cotx\right)=\left(\frac{cosx}{1+sinx}+\frac{sinx}{cosx}\right).\left(\frac{sinx}{1+cosx}+\frac{cosx}{sinx}\right)\)
\(=\left(\frac{cos^2x+sinx.\left(1+sinx\right)}{cosx.\left(1+sinx\right)}\right).\left(\frac{sin^2x+cosx.\left(1+cosx\right)}{sinx.\left(1+cosx\right)}\right)=\left(\frac{cos^2x+sinx+sin^2x}{cosx.\left(1+sinx\right)}\right).\left(\frac{sin^2x+cosx+cos^2x}{sinx.\left(1+cosx\right)}\right)\)
=\(\frac{1}{cosx.sinx}=VP\)
e, \(VT=cos^2x.\left(cos^2x+2sin^2x+sin^2x.tan^2x\right)=cos^2x.\left(1+sin^2x.\left(1+tan^2x\right)\right)=cos^2x.\left(1+tan^2x\right)=cos^2x.\frac{1}{cos^2x}=1=VP\)
c, \(VT=\frac{sin^2x}{cosx.\left(1+tanx\right)}-\frac{cos^2x}{sinx.\left(1+cosx\right)}=\frac{sin^3x.\left(1+cosx\right)-cos^3x.\left(1+tanx\right)}{sinx.cosx.\left(1+tanx\right).\left(1+cosx\right)}\)
=\(\frac{sin^3x+sin^3x.cotx-cos^3x-cos^3.tanx}{\left(sinx+cosx\right)^2}=\frac{sin^3x+sin^2xcosx-cos^3x-cos^2sinx}{\left(sinx+cosx\right)^2}=\frac{sin^2x.\left(sinx+cosx\right)-cos^2x.\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}\)
\(=\frac{\left(sin^2x-cos^2x\right).\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}=\frac{\left(sinx-cosx\right).\left(sinx+cosx\right).\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}=sinx-cosx=VP\)
Đây nha bạn
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 9 2020
a.
\(sinx+cosx+\left(sinx+cosx\right)^2+cos^2x-sin^2x=0\)
\(\Leftrightarrow sinx+cosx+\left(sinx+cosx\right)^2+\left(cosx-sinx\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1+2cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\\1+2cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
29 tháng 4 2019
xét vế phải
( cosa+1-sina)^2
= cos^2 +1+ sin^2+2cosa-2sina-2sinacosa
= 2( 1+ cosa-sina-sinacosa)
= 2( 1-sina) ( 1+cosa)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 5 2020
\(cot^2x-cos^2x=\frac{cos^2x}{sin^2x}-cos^2x=cos^2x\left(\frac{1}{sin^2x}-1\right)=\frac{cos^2x\left(1-sin^2x\right)}{sin^2x}\)
\(=cos^2x.\left(\frac{cos^2x}{sin^2x}\right)=cot^2x.cos^2x\)
\(\frac{cosx+sinx}{cosx-sinx}-\frac{cosx-sinx}{cosx+sinx}=\frac{\left(cosx+sinx\right)^2-\left(cosx-sinx\right)^2}{\left(cosx-sinx\right)\left(cosx+sinx\right)}\)
\(=\frac{cos^2x+sin^2x+2sinx.cosx-\left(cos^2x+sin^2x-2sinx.cosx\right)}{cos^2x-sin^2x}=\frac{4sinx.cosx}{cos2x}=\frac{2sin2x}{cos2x}=2tan2x\)
\(\frac{sin4x+cos2x}{1-cos4x+sin2x}=\frac{2sin2x.cos2x+cos2x}{1-\left(1-2sin^22x\right)+sin2x}=\frac{cos2x\left(2sin2x+1\right)}{sin2x\left(2sin2x+1\right)}=\frac{cos2x}{sin2x}=cot2x\)
\(A=sin^2x\left(sinx+cosx\right)+cos^2x\left(sinx+cosx\right)\)
\(=\left(sin^2x+cos^2x\right)\left(sinx+cosx\right)=sinx+cosx\)
\(B=\frac{sinx}{cosx}\left(\frac{1+cos^2x-sin^2x}{sinx}\right)=\frac{sinx}{cosx}\left(\frac{2cos^2x}{sinx}\right)=2cosx\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 8 2020
c/ ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow\frac{1}{cos^2x}=\frac{1-cos^2x+1-sin^3x}{1-sin^3x}\)
\(\Leftrightarrow\frac{1}{cos^2x}=\frac{sin^2x}{1-sin^3x}+1\)
\(\Leftrightarrow\frac{1}{cos^2x}-1=\frac{sin^2x}{1-sin^3x}\)
\(\Leftrightarrow\frac{1-cos^2x}{cos^2x}=\frac{sin^2x}{1-sin^3x}\)
\(\Leftrightarrow\frac{sin^2x}{cos^2x}=\frac{sin^2x}{1-sin^3x}\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\cos^2x=1-sin^3x\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow1-sin^2x=1-sin^3x\)
\(\Leftrightarrow sin^3x-sin^2x=0\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=1\left(l\right)\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 8 2020
b/ ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\Leftrightarrow\frac{sin2x.sinx+cos2x.cosx}{sinx.cosx}=\frac{sinx}{cosx}-\frac{cosx}{sinx}\)
\(\Leftrightarrow\frac{cos\left(2x-x\right)}{sinx.cosx}=\frac{sin^2x-cos^2x}{sinx.cosx}\)
\(\Leftrightarrow cosx=sin^2x-cos^2x\)
\(\Leftrightarrow cosx=1-2cos^2x\)
\(\Leftrightarrow2cos^2x+cosx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\left(l\right)\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow x=\pm\frac{\pi}{3}+k2\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 4 2019
\(\frac{1-cosx+cos2x}{sin2x-sinx}=\frac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}=\frac{cosx\left(2cosx-1\right)}{sinx\left(2cosx-1\right)}=\frac{cosx}{sinx}=cotx\)
\(A=sin\left(\frac{\pi}{4}+x\right)-sin\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)=sin\left(\frac{\pi}{4}+x\right)-sin\left(\frac{\pi}{4}+x\right)=0\)
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 4 2020
\(\frac{sinx+cotx}{1+sinx.tanx}=\frac{sinx.cosx\left(sinx+cotx\right)}{sinx.cosx\left(1+sinx.tanx\right)}=\frac{cosx\left(sin^2x+cosx\right)}{sinx\left(cosx+sin^2x\right)}=cotx\)
\(\Rightarrow VT=cot^nx\)
\(Vp=\frac{cos^nx.sin^nx\left(sin^nx+cot^nx\right)}{cos^nxsin^nx\left(1+sin^nxcot^nx\right)}=\frac{cos^nx\left(sin^{2n}x+cos^nx\right)}{sin^nx\left(cos^nx+sin^{2n}x\right)}=\frac{cos^nx}{sin^nx}=cot^nx\)
\(\Rightarrow VT=VP\) (đpcm)
3/
\(\frac{sin2x-sinx}{1-cosx+cos2x}=\frac{2sinxcosx-sinx}{1-cosx+2cos^2x-1}=\frac{sinx\left(2cosx-1\right)}{cosx\left(2cosx-1\right)}=\frac{sinx}{cosx}=tanx\)
4/
\(\left(\frac{sinx+cotx}{1+sinx.tanx}\right)^{2014}=\left(\frac{sinx+\frac{1}{tanx}}{1+sinxtanx}\right)^{2014}=\left(\frac{sinxtanx+1}{tanx\left(sinxtanx+1\right)}\right)^{2014}\)
\(=\left(\frac{1}{tanx}\right)^{2014}=cot^{2014}x\)
\(\frac{sin^{2014}x+cot^{2014}x}{1+\left(sinx.tanx\right)^{2014}}=\frac{sin^{2014}x+\frac{1}{tan^{2014}x}}{1+\left(sinx.tanx\right)^{2014}}=\frac{\left(sinxtanx\right)^{2014}+1}{tan^{2014}x\left[\left(sinxtanx\right)^{2014}+1\right]}\)
\(=\frac{1}{tan^{2014}x}=\left(\frac{1}{tanx}\right)^{2014}=cot^{2014}x\)
\(\Rightarrow\left(\frac{sinx+cotx}{1+sinx.tanx}\right)^{2014}=\frac{sin^{2014}x+cot^{2014}x}{1+\left(sinx.tanx\right)^{2014}}\)