tích đúng mình làm cho

mình không hiểu 

sin^2x.sin^2x+cos^2x.cos^2x=1-1/2sin^2(2x)

<=>(1-cos2x)^2/4+(1+cos2x)^2/4=1-1/2sin^2(2x)

<=>(cos^2(2x)+1)/2=1-1/2sin^2(2x)

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

cos^2(2x)+sin^2(2x)=1( luôn đúng)

đpcm

Có: y=sin^4x−cos^4x
        = (sin^2x−cos^2x)(sin^2x+cos^2x)
        = −cos2x
=> −1≤y≤1
=> min y=−1⇔cos2x=1⇔x=kπ
     max y=1⇔cos2x=−1⇔x=π2+kπ
Vậy min y = -1; max y=1

\(y=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x+sin2x\)

\(=1-\dfrac{1}{2}sin^22x+sin2x\)

Đặt \(sin2x=t\in\left[-1;1\right]\Rightarrow y=f\left(t\right)=-\dfrac{1}{2}t^2+t+1\)

\(-\dfrac{b}{2a}=1\) ; \(f\left(-1\right)=-\dfrac{1}{2}\) ; \(f\left(1\right)=\dfrac{3}{2}\)

\(\Rightarrow y_{min}=-\dfrac{1}{2}\) khi \(sin2x=-1\)

\(y_{max}=\dfrac{3}{2}\) khi \(sin2x=1\)

\(\frac{sin2a-2sina}{sin2a+2sina}=\frac{2sina.cosa-2sina}{2sina.cosa+2sina}=\frac{2sina\left(cosa-1\right)}{2sina\left(cosa+1\right)}=\frac{cosa-1}{cosa+1}\)

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

\(\frac{sin^4x-sin^2x+cos^2x}{cos^4x-cos^2x+sin^2x}=\frac{sin^2x\left(sin^2x-1\right)+cos^2x}{cos^2x\left(cos^2x-1\right)+sin^2x}=\frac{-sin^2x.cos^2x+cos^2x}{-cos^2x.sin^2x+sin^2x}\)

\(=\frac{cos^2x\left(1-sin^2x\right)}{sin^2x\left(1-cos^2x\right)}=\frac{cos^4x}{sin^4x}=cot^4x\)

\(\frac{sin^3a-cos^3a}{sina-cosa}=\frac{\left(sina-cosa\right)\left[sin^2a+cos^2a+sina.cosa\right]}{sina-cosa}=1+sina.cosa=1+\frac{1}{2}sin2a\)

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

\(A=\dfrac{sin^2x-cos^2x.\left(1-cos^2x\right)}{cos^2x-sin^2x.\left(1-sin^2x\right)}=\dfrac{sin^2x-cos^2x.sin^2x}{cos^2x-sin^2x.cos^2x}\\ =\dfrac{sin^2x.\left(1-cos^2x\right)}{cos^2x.\left(1-sin^2x\right)}=\dfrac{sin^2x.sin^2x}{cos^2x.cos^2x}=\dfrac{sin^4x}{cos^4x}.\)