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

a: \(=\left(\sin^2\alpha+\cos^2\alpha\right)^2=1^2=1\)

A = 2 ( sin 2 α   +   cos 2 α ) ( sin 4 α   +   cos 4 α   -   sin 2 α cos 2 α )

-   3 ( sin 4 α   +   cos 4 α )

     =   - sin 4 α   -   cos 4 α   -   2 sin 2 α cos 2 α

        =   - ( sin ² α   +   cos ² α ) 2   =   - 1

Ta có:

`sin^4 \alpha + cos^4 \alpha -sin^6 \alpha- cos^6\alpha`

`=sin^4\alpha+cos^4\alpha-(sin^2\alpha+cos^2\alpha)(sin^4\alpha-sin^2\alpha cos^2\alpha+cos^4\alpha)`

`=sin^4\alpha + cos^4\alpha-(sin^4\alpha-sin^2\alpha cos^2\alpha+cos^4\alpha)`

`=sin^2\alpha cos^2\alpha(ĐPCM)`

c: 2(sin^6a+cos^6a)+1

=2[(sin^2a+cos^2a)^3-3*sin^2a*cos^2a]+1

=2-6sin^2acos^2a+1

=3-6*sin^2a*cos^2a

=3(sin^4a+cos^4a)

a:

Sửa đề: =-tana*tanb

 \(VT=\left(\dfrac{sina}{cosa}-\dfrac{sinb}{cosb}\right):\left(\dfrac{cosa}{sina}-\dfrac{cosb}{sinb}\right)\)

\(=\dfrac{sina\cdot cosb-sinb\cdot cosa}{cosa\cdot cosb}:\dfrac{cosa\cdot sinb-cosb\cdot sina}{sina\cdot sinb}\)

\(=\dfrac{sin\left(a-b\right)}{cosa\cdot cosb}\cdot\dfrac{sina\cdot sinb}{sin\left(b-a\right)}\)

\(=-tana\cdot tanb\)

=VP

Vì \(\tan\alpha\cdot\cot\alpha=1\Leftrightarrow\cot\alpha=\dfrac{1}{2,15}=\dfrac{20}{43}\)

\(\sin^2\alpha+\cos^2\alpha=1\\ \Rightarrow\cos^2\alpha=1-0,6^2=0,64\\ \Rightarrow\cos\alpha=0,8=\dfrac{4}{5}\\ \tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{0,6}{0,8}=\dfrac{3}{4}\\ \cot\alpha=\dfrac{1}{\tan\alpha}=\dfrac{1}{0,75}=\dfrac{4}{3}\)

a: \(\cos\alpha=\dfrac{1}{2}\)

\(\tan\alpha=\sqrt{3}\)

\(\cot\alpha=\dfrac{\sqrt{3}}{3}\)