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a) P = sin2α + sin2α.\(\frac{cos\text{α}}{sin\text{α}}\) + cos2α - cos2α.\(\frac{sin\text{α}}{cos\text{α}}\)
=sin2α + sinα.cosα + cos2α - cosα.sinα
=sin2α + cos2α
=1
Vậy P không phụ thuộc vào α
b) Q= -cos4α(2cos2α -1 -2) +sin4α(1 -2sin2α+2)
= -cos4α(cos2α -2) +sin4α(cos2α +2)
=-cos4α.cos2α +2cos4α +sin4α.cos2α +2sin4α
=cos2α(sin4α -cos4α) +2(sin4α +cos4α)
=cos2α [\(\left(\frac{1-cos^22\text{α}}{2}\right)^2-\left(\frac{1+cos^22\text{α}}{2}\right)^2\)]+2.[\(\left(\frac{1-cos^22\text{α}}{2}\right)^2+ \left(\frac{1+cos^22\text{α}}{2}\right)^2\)]
= -cos2α.cos2α +1+cos22α
= -cos22α +1+cos22α
=1
Vậy Q không phụ thuộc vào α
a: \(=1+sin2a+1-sin2a=2\)
b: Sửa đề: \(B=sin^6a+cos^6a+3sin^2acos^2a\)
\(=\left(sin^2a+cos^2a\right)^3-3\cdot sin^2a\cdot cos^2a\cdot\left(sin^2a+cos^2a\right)+3sin^2a\cdot cos^2a\)
=1
\(\sin^4x.\sin^2x+\cos^4x.\cos^2x-\left(\sin^4x+\cos^4x+\dfrac{1}{2}\sin^4x+\dfrac{1}{2}\cos^4x-\dfrac{3}{2}\right)-1=-\sin^4x.\left(1-\sin^2x\right)-cos^4x.\left(1-\cos^2x\right)-\dfrac{1}{2}\left(\sin^4x+\cos^4x\right)+\dfrac{1}{2}=-\left(\sin^4x.\cos^2x+\cos^4x.\sin^2x\right)-\dfrac{1}{2}\left(\left(\sin^2x+\cos^2x\right)^2-2\sin^2x.\cos^2x\right)+\dfrac{1}{2}=-\left(\sin^2x.\cos^2x.\left(\sin^2x+\cos^2x\right)\right)-\dfrac{1}{2}.\left(1-2\sin^2x.\cos^2x\right)+\dfrac{1}{2}=-\sin^2x.\cos^2x+\sin^2x.\cos^2x-\dfrac{1}{2}+\dfrac{1}{2}=0\)
\(6sin^4x-2cos^4x=1\Leftrightarrow6sin^4x-2\left(1-sin^2x\right)^2-1=0\)
\(\Leftrightarrow6sin^4x-2\left(sin^4x-2sin^2x+1\right)-1=0\)
\(\Leftrightarrow4sin^4x+4sin^2x-3=0\)
\(\Leftrightarrow\left(2sin^2x+3\right)\left(2sin^2x-1\right)=0\)
\(\Leftrightarrow2sin^2x=1\Rightarrow sin^2x=\frac{1}{2}\Rightarrow cos^2x=\frac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}sin^4x=\frac{1}{4}\\cos^4x=\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow C=\frac{1}{4}+3.\frac{1}{4}=1\)
a)
\(A=\left(sin\alpha+cos\alpha\right)^2+\left(sin\alpha-cos\alpha\right)^2\)
\(=1+2sin\alpha cos\alpha+1-2sin\alpha cos\alpha=2\) (không phụ thuộc vào \(\alpha\)).
b)
\(B=sin^4\alpha-cos^4\alpha-2sin^2\alpha+1\)
\(=\left(sin^2\alpha+cos^2\alpha\right)\left(sin^2\alpha-cos^2\alpha\right)-2sin^2\alpha+1\)
\(=sin^2\alpha-cos^2\alpha-2sin^2\alpha+1\)
\(=-sin^2\alpha-cos^2\alpha+1\)
\(=-\left(sin^2\alpha+cos^2\alpha\right)+1=-1+1=0\).
a/ Ta có: \(tan\alpha=5\Rightarrow cot\alpha=\frac{1}{5}\) . Đề: \(\frac{sin\alpha}{sin^3\alpha+cos^3\alpha}=\frac{\frac{1}{sin^2\alpha}}{1+\frac{cos^3\alpha}{sin^3\alpha}}=\frac{1+cot^2\alpha}{1+cot^3\alpha}=\frac{1+\left(\frac{1}{5}\right)^2}{1+\left(\frac{1}{5}\right)^3}=\frac{65}{63}\)
b/ Ta có vế trái \(=\frac{sin^2x+cos^2x+cos^2x-sin^2x+\left(sinx+sin3x\right)}{1+2sinx}=\frac{2cos^2x+2.sin2x.cosx}{1+2sinx}=\frac{2cos^2x+4.sinx.cos^2x}{1+2sinx}=\frac{2cos^2x.\left(1+2sinx\right)}{1+2sinx}=2cos^2x\) ( = vế phải)
ta có : \(cos^4\alpha\left(3-2cos^2\alpha\right)+sin^4\alpha\left(3-2sin^2\alpha\right)\)
\(=3cos^4\alpha-2cos^6\alpha+3sin^4\alpha-2sin^6\alpha\)
\(=3\left(sin^4\alpha+cos^4\alpha\right)-2\left(sin^6\alpha+cos^6\alpha\right)\)
\(=3\left(\left(sin^2\alpha+cos^2\alpha\right)-2sin^2\alpha.cos^2\alpha\right)-2\left(\left(sin^2\alpha+cos^2\alpha\right)^3-3sin^2\alpha.cos^2\alpha\left(sin^2+cos^2\alpha\right)\right)\)
\(=3\left(1-2sin^2\alpha.cos^2\alpha\right)-2\left(1-3sin^2\alpha.cos^2\alpha\right)\)
\(=3-6sin^2\alpha.cos^2\alpha-2+6sin^2\alpha.cos^2\alpha=1\) (không phụ thuộc vào \(\alpha\)) (đpcm)