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) \({\sin ^4}\alpha  + {\cos ^4}\alpha  = 1 - 2{\sin ^2}\alpha .{\cos ^2}\alpha \).

\(VT = {\left( {{{\sin }^2}\alpha } \right)^2} + {\left( {{{\cos }^2}\alpha } \right)^2}\)

\(= \left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right) - 2{\sin ^2}\alpha .{\cos ^2}\alpha \)

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

b) \({\sin ^6}\alpha  + {\cos ^6}\alpha  = 1 - 3{\sin ^2}\alpha .{\cos ^2}\alpha \).

\(VT = {\left( {{{\sin }^2}\alpha } \right)^3} + {\left( {{{\cos }^2}\alpha } \right)^3}\)

\(={\left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right)^3} - 3\sin^2 \alpha .\cos^2 \alpha \left( {\sin^2 \alpha  + \cos^2 \alpha } \right)\)

\(= 1 - 3{\sin ^2}\alpha .{\cos ^2}\alpha .1\)

\(= 1 - 3{\sin ^2}\alpha .{\cos ^2}\alpha  = VP\).

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

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

\(= \sqrt {{{\left( {1 - {{\cos }^2}} \right)}^2} + 6{{\cos }^2}\alpha  + 3}  + \sqrt {{{\left( {1 - {{\sin }^2}\alpha } \right)}^2} + 4{{\sin }^2}\alpha } \)

\(= \sqrt {{{\cos }^4}\alpha  + 4{{\cos }^2}\alpha  + 4}  + \sqrt {{{\sin }^4}\alpha  + 2{{\sin }^2}\alpha  + 1} \)

\(= \sqrt {{{\left( {{{\cos }^2}\alpha  + 2} \right)}^2}}  + \sqrt {{{\left( {{{\sin }^2}\alpha  + 1} \right)}^2}} \)

\(= {\cos ^2}\alpha  + 2 + {\sin ^2}\alpha  + 1 = 4 = VP\).