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\(\frac{sin2x-sinx}{1-cosx+cos2x}=\frac{2sinxcosx-sinx}{1-cosx+2cos^2x-1}=\frac{sinx\left(2cosx-1\right)}{cosx\left(2cosx-1\right)}=\frac{sinx}{cosx}=tanx\)
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\(\left(\frac{sinx+cotx}{1+sinx.tanx}\right)^{2014}=\left(\frac{sinx+\frac{1}{tanx}}{1+sinxtanx}\right)^{2014}=\left(\frac{sinxtanx+1}{tanx\left(sinxtanx+1\right)}\right)^{2014}\)
\(=\left(\frac{1}{tanx}\right)^{2014}=cot^{2014}x\)
\(\frac{sin^{2014}x+cot^{2014}x}{1+\left(sinx.tanx\right)^{2014}}=\frac{sin^{2014}x+\frac{1}{tan^{2014}x}}{1+\left(sinx.tanx\right)^{2014}}=\frac{\left(sinxtanx\right)^{2014}+1}{tan^{2014}x\left[\left(sinxtanx\right)^{2014}+1\right]}\)
\(=\frac{1}{tan^{2014}x}=\left(\frac{1}{tanx}\right)^{2014}=cot^{2014}x\)
\(\Rightarrow\left(\frac{sinx+cotx}{1+sinx.tanx}\right)^{2014}=\frac{sin^{2014}x+cot^{2014}x}{1+\left(sinx.tanx\right)^{2014}}\)
\(\frac{sinx+cotx}{1+sinx.tanx}=\frac{sinx.cosx\left(sinx+cotx\right)}{sinx.cosx\left(1+sinx.tanx\right)}=\frac{cosx\left(sin^2x+cosx\right)}{sinx\left(cosx+sin^2x\right)}=cotx\)
\(\Rightarrow VT=cot^nx\)
\(Vp=\frac{cos^nx.sin^nx\left(sin^nx+cot^nx\right)}{cos^nxsin^nx\left(1+sin^nxcot^nx\right)}=\frac{cos^nx\left(sin^{2n}x+cos^nx\right)}{sin^nx\left(cos^nx+sin^{2n}x\right)}=\frac{cos^nx}{sin^nx}=cot^nx\)
\(\Rightarrow VT=VP\) (đpcm)
Giả sử biểu thức xác định
\(\frac{2tanx-sin2x}{\left(sinx+cosx\right)^2-1}=\frac{2tanx-sin2x}{sin^2x+cos^2x+2sinx.cosx-1}=\frac{2tanx-2sinx.cosx}{2sinx.cosx}\)
\(=\frac{sinx}{cosx.sinx.cosx}-1=\frac{1}{cos^2x}-1=\frac{1-cos^2x}{cos^2x}=\frac{sin^2x}{cos^2x}=tan^2x\)
Chứng minh rằng:
\(\left(cos2x-sin2x\right)^2+2\left(sin3x-sinx\right)cosx-1=0\), \(\forall x\in R\)
\(\left(cos2x-sin2x\right)^2+2\left(sin3x-sinx\right).cosx-1\)
\(=2sin^2\left(2x-\frac{\pi}{4}\right)+4cos2x.sinx.cosx-1\)
\(=1-cos\left(4x-\frac{\pi}{2}\right)+2sin2x.cos2x-1\)
\(=-cos\left(\frac{\pi}{2}-4x\right)+sin4x\)
\(=-sin4x+sin4x=0\)
\(\frac{1-sinx-cos2x}{sin2x-cosx}=\frac{1-sinx-\left(1-2sin^2x\right)}{2sinxcosx-cosx}=\frac{2sin^2x-sinx}{2sinxcosx-cosx}\)
\(=\frac{sinx\left(2sinx-1\right)}{cosx\left(2sinx-1\right)}=\frac{sinx}{cosx}=tanx\)
\(2\left[\left(sinx+cosx+1\right)\left(sinx+cosx-1\right)\right]^2\)
\(=2\left[\left(sinx+cosx\right)^2-1\right]^2=2\left(sin^2x+cos^2x+2sinx.cosx-1\right)^2\)
\(=2\left(2sinx.cosx\right)^2=2sin^22x=1-cos4x\)
b/ \(\frac{3-4cos2a+2cos^22a-1}{3+4cos2a+2cos^22a-1}=\frac{2\left(cos^22a-2cos2a+1\right)}{2\left(cos^22a+2cos2a+1\right)}=\frac{\left(cos2a-1\right)^2}{\left(cos2a+1\right)^2}\)
\(\frac{\left(1-2sin^2a-1\right)^2}{\left(2cos^2a-1+1\right)^2}=\frac{4sin^4a}{4cos^4a}=tan^4a\)
c/ \(cos^22x+sin^22x-2sin2x.cos2x+2sin3x.cosx-2sinx.cosx-sin^2x\)
\(=1-sin4x+sin4x+sin2x-sin2x-sin^2x\)
\(=1-sin^2x=cos^2x\)
\(\frac{sin2x-cosx}{2sinx-1}+sinx=\frac{2sinx.cosx-cosx}{2sinx-1}+sinx\)
\(=\frac{cosx\left(2sinx-1\right)}{2sinx-1}+sinx=cosx+sinx=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
\(cot^2x-cos^2x=\frac{cos^2x}{sin^2x}-cos^2x=cos^2x\left(\frac{1}{sin^2x}-1\right)=\frac{cos^2x\left(1-sin^2x\right)}{sin^2x}\)
\(=cos^2x.\left(\frac{cos^2x}{sin^2x}\right)=cot^2x.cos^2x\)
\(\frac{cosx+sinx}{cosx-sinx}-\frac{cosx-sinx}{cosx+sinx}=\frac{\left(cosx+sinx\right)^2-\left(cosx-sinx\right)^2}{\left(cosx-sinx\right)\left(cosx+sinx\right)}\)
\(=\frac{cos^2x+sin^2x+2sinx.cosx-\left(cos^2x+sin^2x-2sinx.cosx\right)}{cos^2x-sin^2x}=\frac{4sinx.cosx}{cos2x}=\frac{2sin2x}{cos2x}=2tan2x\)
\(\frac{sin4x+cos2x}{1-cos4x+sin2x}=\frac{2sin2x.cos2x+cos2x}{1-\left(1-2sin^22x\right)+sin2x}=\frac{cos2x\left(2sin2x+1\right)}{sin2x\left(2sin2x+1\right)}=\frac{cos2x}{sin2x}=cot2x\)
\(A=sin^2x\left(sinx+cosx\right)+cos^2x\left(sinx+cosx\right)\)
\(=\left(sin^2x+cos^2x\right)\left(sinx+cosx\right)=sinx+cosx\)
\(B=\frac{sinx}{cosx}\left(\frac{1+cos^2x-sin^2x}{sinx}\right)=\frac{sinx}{cosx}\left(\frac{2cos^2x}{sinx}\right)=2cosx\)
Ủa câu này nãy làm rồi mà bạn chưa hiểu hay sao?
\(VT=cos^22x+sin^22x-2sin2x.cos2x+2sin3x.cosx-2sinx.cosx-sin^{2x}\)
Ở đây ta lần lượt có:
\(cos^22x+sin^22x=1\)
\(2sin2x.cos2x=sin4x\)
\(2sin3x.cosx=sin4x+sin2x\)
\(2sinx.cosx=sin2x\)
Ghép lại sẽ được:
\(VT=1-sin4x+sin4x+sin2x-sin2x-sin^2x=1-sin^2x=cos^2x\)