Bài 23 trang 201 Đại số 10 Nâng cao: Chứng minh các biểu thức sau không phụ thuộc vào α...

Chứng minh các biểu thức sau không phụ thuộc vào α

a) $\sqrt {{{\sin }^4}\alpha  + 4(1 - {{\sin }^2}\alpha )}  + \sqrt {{{\cos }^4}\alpha  + 4{{\sin }^2}\alpha }$

b) $2(si{n^6}\alpha {\rm{ }} + {\rm{ }}co{s^6}\alpha {\rm{ }}){\rm{ }}-{\rm{ }}3(co{s^4}\alpha {\rm{ }} + {\rm{ }}si{n^4}\alpha {\rm{ }})$

c) ${2 \over {\tan \alpha  - 1}} + {{\cot \alpha  + 1} \over {\cot \alpha  - 1}}\,\,\,\,(\tan \alpha  \ne 1)$

Đáp án

a) Ta có:

$\eqalign{ & \sqrt {{{\sin }^4}\alpha + 4(1 - {{\sin }^2}\alpha )} = \sqrt {{{(2 - {{\sin }^2}\alpha )}^2}} \cr & = 2 - {\sin ^2}\alpha \,\,\,\,\,\,\,\,\,({\sin ^2}\alpha \le 1) \cr & \sqrt {{{\cos }^4}\alpha + 4(1 - {{\cos }^2})} = \sqrt {{{(2 - {{\cos }^2}\alpha )}^2}} \cr & = 2 - {\cos ^2}\alpha \,\,\,\,\,\,\,\,(co{s^2}\alpha \le1) \cr}$

Vậy $\sqrt {{{\sin }^4}\alpha  + 4(1 - {{\sin }^2}\alpha )}  + \sqrt {{{\cos }^4}\alpha  + 4{{\sin }^2}\alpha }$

$= {\rm{ }}4{\rm{ }}-{\rm{ }}si{n^2}\alpha {\rm{ }}-{\rm{ }}co{s^2}\alpha {\rm{ }} = {\rm{ }}4{\rm{ }}-{\rm{ }}1 = {\rm{ }}3$

b) Ta có:

$si{n^6}\alpha {\rm{ }} + {\rm{ }}co{s^6}\alpha$ 

$= {\rm{ }}(si{n^2}\alpha {\rm{ }} + {\rm{ }}co{s^2}\alpha ){\rm{ }}-{\rm{ }}3si{n^2}\alpha co{s^2}\alpha {\rm{ }}(si{n^2}\alpha {\rm{ }} + {\rm{ }}co{s^2}\alpha )$

$= {\rm{ }}1{\rm{ }}-{\rm{ }}3si{n^2}\alpha {\rm{ }}co{s^2}\alpha$

$co{s^4}\alpha {\rm{ }} + {\rm{ }}si{n^4}\alpha {\rm{ }} = {\rm{ }}{(co{s^2}\alpha {\rm{ }} + {\rm{ }}si{n^2}\alpha )^2}-{\rm{ }}2si{n^2}\alpha {\rm{ }}co{s^2}\alpha$

$= {\rm{ }}1{\rm{ }} - {\rm{ }}2si{n^2}\alpha {\rm{ }}co{s^2}\alpha$

Suy ra: 

$\eqalign{ & 2\left( {{{\sin }^6}\alpha + {{\cos }^6}\alpha } \right) - 3({\cos ^4}\alpha + {\sin ^4}\alpha ) \cr & = 2 - 6{\sin ^2}\alpha {\cos ^2}\alpha - 3(1 - 2{\sin ^2}\alpha {\cos ^2}\alpha ) \cr & = 2 - 3 = - 1 \cr}$

c) Ta có:

$\eqalign{ & {2 \over {\tan \alpha - 1}} + {{\cot \alpha + 1} \over {\cot \alpha - 1}}\,\,\,\, \cr&= {2 \over {{1 \over {\cot \alpha }} - 1}} + {{\cos \alpha + 1} \over {\cot \alpha - 1}} \cr & = {{2\cot \alpha } \over {1 - \cot \alpha }} + {{\cot \alpha + 1} \over {\cot \alpha - 1}} = {{\cot \alpha - 1} \over {1 - \cot \alpha }} = - 1 \cr}$