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\(A=\left(sin^2a+cos^2a\right)\left(sin^4a-sin^2acos^2a+cos^4a\right)+3sin^2acos^2a\)
A = \(sin^4+2sin^2acos^2a+cos^4a=\left(sin^2a+cos^2a\right)^2=1\)
\(\tan\alpha+\cot\alpha=3\)
\(\Leftrightarrow\frac{\sin\alpha}{\cos\alpha}+\frac{\cos\alpha}{\sin\alpha}=3\)
\(\Leftrightarrow\frac{\sin^2\alpha+\cos^2\alpha}{\sin\alpha.\cos\alpha}=3\)
\(\Leftrightarrow\frac{1}{\sin\alpha.\cos\alpha}=3\)
\(\Rightarrow\sin\alpha.\cos\alpha=\frac{1}{3}\)
\(\frac{1-tana}{1+tana}=\frac{1-\frac{sina}{cosa}}{1+\frac{sina}{cosa}}=\frac{\frac{1}{cosa}\left(cosa-sina\right)}{\frac{1}{cosa}\left(cosa+sina\right)}=\frac{cosa-sina}{cosa+sina}\)
A = \(\left(sin^2a+cos^2a\right)^2=1^2=1\)
D = \(sin^2\left(sin^2B+cos^2B\right)+cos^2a=sin^2a+cos^2a=1\)
\(A=\left(\sin\alpha+\cos\alpha+\sin\alpha-\cos\alpha\right)^2-2\left(\sin\alpha+\cos\alpha\right)\left(\sin\alpha-\cos\alpha\right)\)
\(=4\sin^2\alpha-2\sin^2\alpha+2\cos^2\alpha=2\left(\sin^2\alpha+\cos^2\alpha\right)=2\)
\(B=\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)=\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\)
\(=\left(\sin^2\alpha+\cos^2\alpha\right)^2-1=0\)
\(C=3\left(\sin^4\alpha+\cos^4\alpha\right)-2\sin^2\alpha.\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)=3\left(\sin^4\alpha+\cos^4\alpha\right)-2\sin^2\alpha.\cos^2\alpha\)
\(=3\left(\sin^2\alpha+\cos^2\alpha-\frac{1}{9}\right)^2-\frac{1}{9}=\frac{61}{27}\)
bài 1: ta có : \(cos^220+cos^240+cos^250+cos^270\)
\(=cos^220+cos^270+cos^240+cos^250\)
\(=cos^220+cos^2\left(90-20\right)+cos^240+cos^2\left(90-40\right)\)
\(=cos^220+sin^220+cos^240+sin^240=1+1=2\)
bài 2: a) ta có : \(cot^2\alpha-cos^2\alpha=cos^2\alpha\left(\dfrac{1}{sin^2\alpha}-1\right)=cos^2\alpha.\left(\dfrac{1-sin^2\alpha}{sin^2\alpha}\right)\)
\(=cos^2\alpha.\left(\dfrac{cos^2\alpha}{sin^2\alpha}\right)=cos^2\alpha.cot^2\alpha\left(đpcm\right)\)
b) ta có : \(sin^2\alpha+cos^2\alpha=1\Leftrightarrow sin^2\alpha=1-cos^2\alpha\)
\(\Leftrightarrow sin^2\alpha=\left(1-cos\alpha\right)\left(1+cos\alpha\right)\Leftrightarrow\dfrac{1+cos\alpha}{sin\alpha}=\dfrac{sin\alpha}{1-cos\alpha}\left(đpcm\right)\)
Ta có:
= sin 6 α + cos 6 α + 3 sin 2 α . cos 2 α . ( sin 2 α + cos 2 α ) v ì sin 2 α + cos 2 α = 1
= ( sin 2 α ) 3 + 3 sin 2 α 2 cos 2 + 3 sin 2 α . cos 2 α 2 + cos 2 α 3
Đáp án cần chọn là: B