Giải bài 3.5 trang 33 sách bài tập toán 10 - Kết nối tri thức với cuộc sống

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) \(\sqrt {{{\sin }^4}\alpha + 6{{\cos }^2}\alpha + 3} + \sqrt {{{\cos }^4}\alpha + 4{{\sin }^2}\alpha } = 4.\)

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