Bài 35 trang 197 Sách bài tập (SBT) Toán Đại số 10

Chứng minh rằng các biểu thức sau là những số không phụ thuộc \(\alpha \)

a) \(A = 2({\sin ^6}\alpha  + c{\rm{o}}{{\rm{s}}^6}\alpha ) - 3({\sin ^4}\alpha  + c{\rm{o}}{{\rm{s}}^4}\alpha )\)

b) \(A = 4({\sin ^4}\alpha  + c{\rm{o}}{{\rm{s}}^4}\alpha ) - c{\rm{os4}}\alpha \)

c) \(C = 8(c{\rm{o}}{{\rm{s}}^8}\alpha  - {\sin ^8}\alpha ) - \cos 6\alpha  - 7\cos 2\alpha \)

Gợi ý làm bài

a) \(A = 2({\sin ^2}\alpha  + c{\rm{o}}{{\rm{s}}^2}\alpha )({\sin ^4}\alpha  + c{\rm{o}}{{\rm{s}}^4}\alpha  - {\sin ^2}\alpha co{s^2}\alpha ) - 3({\sin ^4}\alpha  + c{\rm{o}}{{\rm{s}}^4}\alpha )\)

= \( - {\sin ^4}\alpha  - {\cos ^4}\alpha  - 2{\sin ^2}{\cos ^2}\alpha \)

= \( - {({\sin ^2}\alpha  + {\cos ^2}\alpha )^2} =  - 1\)

b) \(A = 4{\rm{[}}{({\sin ^2}\alpha  + c{\rm{o}}{{\rm{s}}^2}\alpha )^2} - 2{\sin ^2}\alpha c{\rm{o}}{{\rm{s}}^2}\alpha {\rm{]}} - c{\rm{os4}}\alpha \)

= \(4\left( {1 - {1 \over 2}{{\sin }^2}2\alpha } \right) - 1 + 2{\sin ^2}2\alpha  = 3\)

c) \(C = 8(c{\rm{o}}{{\rm{s}}^4}\alpha  - {\sin ^4}\alpha )(c{\rm{o}}{{\rm{s}}^4}\alpha  + {\sin ^4}\alpha ) - \cos 6\alpha  - 7\cos 2\alpha \)

\( = 8(c{\rm{o}}{{\rm{s}}^2}\alpha  - {\sin ^2}\alpha )(c{\rm{o}}{{\rm{s}}^2}\alpha  + {\sin ^2}\alpha ){\rm{[}}{(c{\rm{o}}{{\rm{s}}^2}\alpha  + {\sin ^2}\alpha )^2} - 2{\sin ^2}\alpha c{\rm{o}}{{\rm{s}}^2}\alpha {\rm{]}} - \cos 6\alpha  - 7\cos 2\alpha \)

\( = 8c{\rm{os}}2\alpha \left( {1 - {1 \over 2}si{n^2}2\alpha } \right) - c{\rm{os6}}\alpha {\rm{ - 7cos2}}\alpha \)

\( = c{\rm{os}}2\alpha  - 4\cos 2\alpha si{n^2}2\alpha  - c{\rm{os(4}}\alpha  + {\rm{2}}\alpha )\)

\( = c{\rm{os}}2\alpha  - 2\sin 4\alpha sin2\alpha  - c{\rm{os4}}\alpha c{\rm{os2}}\alpha  + \sin 4\alpha sin2\alpha \)

\( = c{\rm{os}}2\alpha  - (\cos 4\alpha \cos 2\alpha  + \sin {\rm{4}}\alpha \sin {\rm{2}}\alpha )\)

\( = \cos 2\alpha  - c{\rm{os2}}\alpha {\rm{ = 0}}\)

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