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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) \(C = 8(c{\rm{o}}{{\rm{s}}^4}\alpha – {\sin ^4}\alpha )(c{\rm{o}}{{\rm{s}}^4}\alpha + {\sin ^4}\alpha ) – \cos 6\alpha – 7\cos 2\alpha \)
\( = 8(c{\rm{o}}{{\rm{s}}^2}\alpha – {\sin ^2}\alpha )(c{\rm{o}}{{\rm{s}}^2}\alpha + {\sin ^2}\alpha ){\rm{[}}{(c{\rm{o}}{{\rm{s}}^2}\alpha + {\sin ^2}\alpha )^2} – 2{\sin ^2}\alpha c{\rm{o}}{{\rm{s}}^2}\alpha {\rm{]}} – \cos 6\alpha – 7\cos 2\alpha \)
\( = 8c{\rm{os}}2\alpha \left( {1 – {1 \over 2}si{n^2}2\alpha } \right) – c{\rm{os6}}\alpha {\rm{ – 7cos2}}\alpha \)
\( = c{\rm{os}}2\alpha – 4\cos 2\alpha si{n^2}2\alpha – c{\rm{os(4}}\alpha + {\rm{2}}\alpha )\)
\( = c{\rm{os}}2\alpha – 2\sin 4\alpha sin2\alpha – c{\rm{os4}}\alpha c{\rm{os2}}\alpha + \sin 4\alpha sin2\alpha \)
\( = c{\rm{os}}2\alpha – (\cos 4\alpha \cos 2\alpha + \sin {\rm{4}}\alpha \sin {\rm{2}}\alpha )\)
\( = \cos 2\alpha – c{\rm{os2}}\alpha {\rm{ = 0}}\)