Tại sao sin^4x+cos^4x=1-1/2.sin^2(2x) và sin^6x+cos^6x=1-3/4.sin^2(2x) ạ?

\[\begin{array}{l} {\sin ^4}x + {\cos ^4}x = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x\\ = 1 - 2{\sin ^2}x{\cos ^2}x = 1 - \frac{1}{2}.\left( {4{{\sin }^2}x{{\cos }^2}x} \right) = 1 - \frac{1}{2}{\sin ^2}2x\\ {\sin ^6}x + {\cos ^6}x = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^3} - 3{\sin ^2}x{\cos ^2}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\\ = 1 - 3{\sin ^2}x{\cos ^2}x = 1 - \frac{3}{4}.\left( {4{{\sin }^2}x{{\cos }^2}x} \right) = 1 - \frac{3}{4}{\sin ^2}2x \end{array}\]