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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!
a/ \(A=\frac{cot^2a-cos^2a}{cot^2a}-\frac{sina.cosa}{cota}\)
\(=\frac{\frac{cos^2a}{sin^2a}-cos^2a}{\frac{cos^2a}{sin^2a}}-\frac{sina.cosa}{\frac{cosa}{sina}}\)
\(=\left(1-sin^2a\right)-sin^2a=1\)
b/ \(B=\left(cosa-sina\right)^2+\left(cosa+sina\right)^2+cos^4a-sin^4a-2cos^2a\)
\(=cos^2a-2cosa.sina+sin^2a+cos^2a+2cosa.sina+sin^2a+\left(cos^2a+sin^2a\right)\left(cos^2a-sin^2a\right)-2cos^2a\)
\(=2+\left(cos^2a-sin^2a\right)-2cos^2a\)
\(=2-sin^2a-cos^2a=2-1=1\)
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 :
\(B=sin^6\alpha+cos^6\alpha+3sin^2\alpha.cos^2\alpha\)
\(=\left(sin^2\alpha+cos^2\alpha\right)^3-3sin^2\alpha.cos^2\alpha\left(sin^2\alpha+cos^2\alpha\right)+3sin^2\alpha.cos^2\alpha\)
\(=1-3sin^2\alpha.cos^2\alpha+3sin^2\alpha.cos^2\alpha=1\)
Bài 2:
a: \(\sin a=\sqrt{1-\left(\dfrac{4}{5}\right)^2}=\dfrac{3}{5}\)
\(P=4\cdot\sin^2a-6\cdot\cos^2a\)
\(=4\cdot\dfrac{9}{25}-6\cdot\dfrac{16}{25}\)
\(=\dfrac{36-64}{25}=\dfrac{-28}{25}\)
b: \(A=\sin^6a+\cos^6a+3\cdot\sin^2a\cdot\cos^2a\)
\(=\left(\sin^2a+\cos^2a\right)^3-3\sin^2a\cdot\cos^2a\cdot\left(\sin^2a+\cos^2a\right)+3\cdot\sin^2a\cdot\cos^2a\)
\(=1-3\sin^2a\cdot\cos^2a+3\sin^2a\cdot\cos^2a\)
=1