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\(A=sin^6\alpha+cos^6\alpha+3sin^2\alpha-cos^2\alpha\)
\(=\left(sin^2\alpha\right)^3+\left(cos^2\alpha\right)^3+3sin^2\alpha-cos^2\alpha\)
\(=\left(sin^2\alpha+cos^2\alpha\right)\left(sin^4\alpha+cos^4\alpha-sin^2\alpha.cos^2\alpha\right)+3sin^2\alpha-cos^2\alpha\)
\(=sin^4\alpha+cos^4\alpha-sin^2\alpha.cos^2\alpha+3sin^2\alpha-cos^2\alpha\)
\(=\left(sin^2\alpha+cos^2\alpha\right)^2-2sin^2\alpha.cos^2\alpha-sin^2\alpha.cos^2\alpha+3sin^2\alpha-cos^2\alpha\)
\(1-3sin^2\alpha.cos^2\alpha+3sin^2\alpha-cos^2\alpha=3sin^2\alpha\left(1-cos^2\alpha\right)+\left(1-cos^2\alpha\right)\)
\(=\left(3sin^2\alpha+1\right).sin^2\alpha=0\)
sữa đề chút nha :
+) ta có : \(A=\dfrac{1+2sin\alpha.cos\alpha}{cos^2\alpha-sin^2\alpha}=\dfrac{\left(sin\alpha+cos\alpha\right)^2}{\left(sin\alpha+cos\alpha\right)\left(cos\alpha-sin\alpha\right)}=\dfrac{sin\alpha+cos\alpha}{cos\alpha-sin\alpha}\)
+) ta có :
\(B=sin^6\alpha+cos^6\alpha+3sin^2\alpha.cos^2\alpha\)
\(=\left(sin^2\alpha+cos^2\alpha\right)^3-3sin^2\alpha.cos^2\alpha\left(sin^2\alpha+cos^2\alpha\right)+3sin^2\alpha.cos^2\alpha\)
\(=1-3sin^2\alpha.cos^2\alpha+3sin^2\alpha.cos^2\alpha=1\)
\(sin^4a+cos^4a+2sin^2a.cos^2a=\left(sin^2a+cos^2a\right)^2=1^2=1\)
b) \(sin^6a+cos^6a+3sin^2a.cos^2a=\left(sin^2a+cos^2a\right)\left(sin^4a-sin^2a.cos^2a+cos^4a\right)+3sin^2a.cos^2a=sin^4a+2sin^2a.cos^2a+cos^4a=\left(sin^2a+cos^2a\right)^2=1\)
Đặt \(\sin^2\alpha=x\Rightarrow\cos^2\alpha=1-\sin^2\alpha\)
\(A=x^3+\left(1-x\right)^3+3x-\left(1-x\right)=x^3+1-3x+3x^2-x^3+3x-1+x=3x^2+x\)
Vậy \(A=3\sin^4\alpha+\sin^2\alpha\). NHỚ NHA!