Đề bài

Chứng minh rằng: a) \({\sin ^4}\alpha  + {\cos ^4}\alpha  = 1 - 2{\sin ^2}\alpha .{\cos ^2}\alpha .\) b) \({\sin ^6}\alpha  + {\cos ^6}\alpha  = 1 - 3{\sin ^2}\alpha .{\cos ^2}\alpha .\) c) \(\sqrt {{{\sin }^4}\alpha  + 6{{\cos }^2}\alpha  + 3}  + \sqrt {{{\cos }^4}\alpha  + 4{{\sin }^2}\alpha }  = 4.\)

Lời giải chi tiết

a) \({\sin ^4}\alpha  + {\cos ^4}\alpha  = 1 - 2{\sin ^2}\alpha .{\cos ^2}\alpha .\)\(\begin{array}{l}VT = {\left( {{{\sin }^2}\alpha } \right)^2} + {\left( {{{\cos }^2}\alpha } \right)^2}\\ = \left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right) - 2{\sin ^2}\alpha .{\cos ^2}\alpha  = 1 - 2{\sin ^2}\alpha .{\cos ^2}\alpha  = VP\end{array}\)b) \({\sin ^6}\alpha  + {\cos ^6}\alpha  = 1 - 3{\sin ^2}\alpha .{\cos ^2}\alpha .\)\(\begin{array}{l}VT = {\left( {{{\sin }^2}\alpha } \right)^3} + {\left( {{{\cos }^2}\alpha } \right)^3}\\ = {\left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right)^3} - 3\sin \alpha .\cos \alpha \left( {\sin \alpha  + \cos \alpha } \right)\\ = 1 - 3{\sin ^2}\alpha .{\cos ^2}\alpha \left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right)\\ = 1 - 3{\sin ^2}\alpha .{\cos ^2}\alpha  = VP\end{array}\)c) \(\sqrt {{{\sin }^4}\alpha  + 6{{\cos }^2}\alpha  + 3}  + \sqrt {{{\cos }^4}\alpha  + 4{{\sin }^2}\alpha }  = 4.\)

\(\begin{array}{l}VT = \sqrt {{{\left( {{{\sin }^2}\alpha } \right)}^2} + 6{{\cos }^2}\alpha  + 3}  + \sqrt {{{\left( {{{\cos }^2}\alpha } \right)}^2} + 4{{\sin }^2}\alpha } \\ = \sqrt {{{\left( {1 - {{\cos }^2}} \right)}^2} + 6{{\cos }^2}\alpha  + 3}  + \sqrt {{{\left( {1 - {{\sin }^2}\alpha } \right)}^2} + 4{{\sin }^2}\alpha } \\ = \sqrt {{{\cos }^4}\alpha  + 4{{\cos }^2}\alpha  + 4}  + \sqrt {{{\sin }^4}\alpha  + 2{{\sin }^2}\alpha  + 1} \\ = \sqrt {{{\left( {{{\cos }^2}\alpha  + 2} \right)}^2}}  + \sqrt {{{\left( {{{\sin }^2}\alpha  + 1} \right)}^2}} \\ = {\cos ^2}\alpha  + 2 + {\sin ^2}\alpha  + 1 = 4 = VP\end{array}\)