Toán Tại sao sin^4x+cos^4x=1-1/2.sin^2(2x) và sin^6x+cos^6x=1-3/4.sin^2(2x) ạ? 12/09/2021 By Valentina Tại sao sin^4x+cos^4x=1-1/2.sin^2(2x) và sin^6x+cos^6x=1-3/4.sin^2(2x) ạ?
\[\begin{array}{l} {\sin ^4}x + {\cos ^4}x = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} – 2{\sin ^2}x{\cos ^2}x\\ = 1 – 2{\sin ^2}x{\cos ^2}x = 1 – \frac{1}{2}.\left( {4{{\sin }^2}x{{\cos }^2}x} \right) = 1 – \frac{1}{2}{\sin ^2}2x\\ {\sin ^6}x + {\cos ^6}x = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^3} – 3{\sin ^2}x{\cos ^2}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\\ = 1 – 3{\sin ^2}x{\cos ^2}x = 1 – \frac{3}{4}.\left( {4{{\sin }^2}x{{\cos }^2}x} \right) = 1 – \frac{3}{4}{\sin ^2}2x \end{array}\] Trả lời
\[\begin{array}{l}
{\sin ^4}x + {\cos ^4}x = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} – 2{\sin ^2}x{\cos ^2}x\\
= 1 – 2{\sin ^2}x{\cos ^2}x = 1 – \frac{1}{2}.\left( {4{{\sin }^2}x{{\cos }^2}x} \right) = 1 – \frac{1}{2}{\sin ^2}2x\\
{\sin ^6}x + {\cos ^6}x = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^3} – 3{\sin ^2}x{\cos ^2}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\\
= 1 – 3{\sin ^2}x{\cos ^2}x = 1 – \frac{3}{4}.\left( {4{{\sin }^2}x{{\cos }^2}x} \right) = 1 – \frac{3}{4}{\sin ^2}2x
\end{array}\]