\(A=2sin2x.cos2x.cos4x=sin4x.cos4x=\frac{1}{2}sin8x\)

\(B=sin^4x+cos^6x-6sin^2x.cos^2x\)

\(=\left(sin^2x+cos^2x\right)^2-8sin^2x.cos^2x\)

\(=1-2\left(2sinx.cosx\right)^2=1-2sin^22x=cos4x\)

\(C=\frac{cos2a+1-2cos^22a}{2sin2a.cos2a+sin2a}=\frac{\left(1-cos2a\right)\left(2cos2a+1\right)}{sin2a\left(2cos2a+1\right)}=\frac{1-cos2a}{sin2a}\)

\(=\frac{1-\left(1-2sin^2a\right)}{2sina.cosa}=\frac{2sin^2a}{2sina.cosa}=\frac{sina}{cosa}=tana\)

\(D=\frac{2cos3a.cos2a+cos3a}{2sin3a.cos2a+sin3a}=\frac{cos3a\left(2cos2a+1\right)}{sin3a\left(2cos2a+1\right)}=\frac{cos3a}{sin3a}=cot3a\)

\(E=\frac{1}{2}-\frac{1}{2}cos\left(\frac{\pi}{4}+x\right)-\frac{1}{2}+\frac{1}{2}cos\left(\frac{\pi}{4}+x\right)\)

\(=\frac{1}{2}\left[cos\left(\frac{\pi}{4}+x\right)-cos\left(\frac{\pi}{4}-x\right)\right]=-sin\frac{\pi}{4}.sinx=-\frac{\sqrt{2}}{2}sinx\)

\(A=\frac{\left(1+cos2x\right)}{cos2x}.tanx=\frac{\left(1+2cos^2x-1\right)}{cos2x}.\frac{sinx}{cosx}=\frac{2cos^2x.sinx}{cos2x.cosx}=\frac{2sinx.cosx}{cos2x}=\frac{sin2x}{cos2x}=tan2x\)

\(B=\frac{1+2sin2a.cos2a-1+2sin^22a}{1+2sin2a.cos2a+2cos^22a-1}=\frac{2sin2a\left(sin2a+cos2a\right)}{2cos2a\left(sin2a+cos2a\right)}=\frac{sin2a}{cos2a}=tan2a\)

\(C=\frac{2sina.cosa+sina}{1+2cos^2a-1+cosa}=\frac{sina\left(2cosa+1\right)}{cosa\left(2cosa+1\right)}=\frac{sina}{cosa}=tana\)

Quên cách giải ptlg rồi nên lm câu 4 =.=

\(\cos3x=\cos\left(2x+x\right)=\cos2x.\cos x-\sin2x.\sin x\)

\(=\left(2\cos^2x-1\right)\cos x-2\sin^2x.\cos x\)

\(=2\cos^3x-\cos x-2\sin^2x.\cos x\)

\(\Rightarrow A=\frac{1+\cos x+2\cos^2x-1+2\cos^3x-\cos x-2\sin^2x.\cos x}{2\cos^2x-1+\cos x}\)

\(=\frac{2\cos^2x+2\cos^3x-2\sin^2x.\cos x}{2\cos^2x-1+\cos x}\)

\(=\frac{2\cos^2x+2\cos^3x-2\left(1-\cos^2x\right).\cos x}{2\cos^2x-1+\cos x}\)

\(=\frac{2\cos^2x+2\cos^3x-2\cos x+2\cos^3x}{2\cos^2x-1+\cos x}\)

\(=\frac{2\cos x\left(2\cos^2x+\cos x-1\right)}{2\cos^2x-1+\cos x}=2\cos x\)

\(\dfrac{1+cosx+cos2x+cos3x}{2cos^2x+cosx-1}=\dfrac{1+cosx+2cos^2x-1+4cos^3x-3cosx}{2cos^2x+cosx-1}\)

\(=\dfrac{4cos^3x+2cos^2x-2cosx}{2cos^2x+cosx-1}=\dfrac{2cosx\left(2cos^2x+cosx-1\right)}{2cos^2x+cosx-1}=2cosx\)

\(2\left[\left(sinx+cosx+1\right)\left(sinx+cosx-1\right)\right]^2\)

\(=2\left[\left(sinx+cosx\right)^2-1\right]^2=2\left(sin^2x+cos^2x+2sinx.cosx-1\right)^2\)

\(=2\left(2sinx.cosx\right)^2=2sin^22x=1-cos4x\)

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/ \(cos^22x+sin^22x-2sin2x.cos2x+2sin3x.cosx-2sinx.cosx-sin^2x\)

\(=1-sin4x+sin4x+sin2x-sin2x-sin^2x\)

\(=1-sin^2x=cos^2x\)

\(\frac{1+sin^2x}{1-sin^2x}=\frac{1+sin^2x}{cos^2x}=\frac{1}{cos^2x}+\frac{sin^2x}{cos^2x}=1+tan^2x+tan^2x=1+2tan^2x\)

\(\frac{sin^3a-cos^3a}{sina-cosa}-sina.cosa=\frac{\left(sina-cosa\right)\left(sin^2a+cos^2a+sina.cosa\right)}{sina-cosa}-sina.cosa\)

\(=sin^2a+cos^2a+sina.cosa-sina.cosa=1\)

\(\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cosx.cos2x}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)

\(\frac{1-2sin^2a}{cosa+sina}+\frac{2cos^2a-1}{cosa-sina}=\frac{cos^2a-sin^2a}{cosa+sina}+\frac{cos^2a-sin^2a}{cosa-sina}\)

\(=\frac{\left(cosa+sina\right)\left(cosa-sina\right)}{cosa+sina}+\frac{\left(cosa+sina\right)\left(cosa-sina\right)}{cosa-sina}=cosa-sina+cosa+sina=2cosa\)

\(\frac{1-cosx+cos2x}{sin2x-sinx}=\frac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}=\frac{cosx\left(2cosx-1\right)}{sinx\left(2cosx-1\right)}=\frac{cosx}{sinx}=cotx\)