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:
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\(si{n^6}\alpha {\rm{ }} + {\rm{ }}co{s^6}\alpha \)
\( = {\rm{ }}(si{n^2}\alpha {\rm{ }} + {\rm{ }}co{s^2}\alpha ){\rm{ }}-{\rm{ }}3si{n^2}\alpha co{s^2}\alpha {\rm{ }}(si{n^2}\alpha {\rm{ }} + {\rm{ }}co{s^2}\alpha )\)
\( = {\rm{ }}1{\rm{ }}-{\rm{ }}3si{n^2}\alpha {\rm{ }}co{s^2}\alpha \)
\(co{s^4}\alpha {\rm{ }} + {\rm{ }}si{n^4}\alpha {\rm{ }} = {\rm{ }}{(co{s^2}\alpha {\rm{ }} + {\rm{ }}si{n^2}\alpha )^2}-{\rm{ }}2si{n^2}\alpha {\rm{ }}co{s^2}\alpha \)
\( = {\rm{ }}1{\rm{ }} - {\rm{ }}2si{n^2}\alpha {\rm{ }}co{s^2}\alpha \)
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:
\(\eqalign{
& {2 \over {\tan \alpha - 1}} + {{\cot \alpha + 1} \over {\cot \alpha - 1}}\,\,\,\, \cr&= {2 \over {{1 \over {\cot \alpha }} - 1}} + {{\cos \alpha + 1} \over {\cot \alpha - 1}} \cr
& = {{2\cot \alpha } \over {1 - \cot \alpha }} + {{\cot \alpha + 1} \over {\cot \alpha - 1}} = {{\cot \alpha - 1} \over {1 - \cot \alpha }} = - 1 \cr} \)