Question

The maximum value of $y=\frac{1}{{\sin }^{6}x+{\cos }^{6}x}$ is

Answer 1

$y=\frac{1}{{\sin }^{6}x+{\cos }^{6}x}Weknowthat{a}^3+b^3={(a+b)}^3-3ab(a+b)AndSomeImpResultsare{\sin }^{2}x+{\cos }^{2}x=1and2\sin x\cos x=\sin 2xIfa={\sin }^{2}xandb={\cos }^{2}xThen{\sin }^{6}x+{\cos }^{6}x=({\sin }^{2}x+{\cos }^{2}x{)}^3-3{\sin }^{2}x{\cos }^{2}x({\sin }^{2}x+{\cos }^{2}x)Sowehave{\sin }^{6}x+{\cos }^{6}x=1-3{\sin }^{2}x{\cos }^{2}x{\sin }^{6}x+{\cos }^{6}x=1-\frac{3}{4}{\sin }^{2}2xForytobemaximum{\sin }^{6}x+{\cos }^{6}xshouldbeminimumSo{\sin }^{2}2x=1y_{max}=\frac{1}{1-\frac{3}{4}}=4$