The maximum value of
y = 1/ (sin^6x + cos^6x) is

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Solution

$C o n s i d e r \sin^{6} x + \cos^{6} x = {(\sin^{2} x)}^{3} + {(\cos^{2} x)}^{3} U S e a^{3} + b^{3} = (a + b) (a^{2} + b^{2} - a b) = (\sin^{2} x + \cos^{2} x) (\sin^{4} x + \cos^{4} x - \sin^{2} x \cos^{2}) = (\sin^{4} x + \cos^{4} x - \sin^{2} x \cos^{2}) = ({(\sin^{2} x)}^{2} + {(\cos^{2} x)}^{2} - \sin^{2} x \cos^{2}) = ({(\sin^{2} x)}^{2} + {(\cos^{2} x)}^{2} + 2 \sin^{2} x \cos^{2} - 3 \sin^{2} x \cos^{2}) = ({(\sin^{2} x + \cos^{2} x)}^{2} - 3 \sin^{2} x \cos^{2}) = (1^{2} - 3 \sin^{2} x \cos^{2}) = (1 - \frac{3}{4} {(2 \sin x \cos x)}^{2}) = (1 - \frac{3}{4} \sin^{2} 2 x) S q u a r e o f a r e a l n u m b e r i s n o n - n e g a t i v e \sin^{2} 2 x \geq 0 A l s o, w e k n o w \sin x \leq 1 0 \leq \sin^{2} 2 x \leq 1 0 \geq - \sin^{2} 2 x \geq - 1 0 \geq - \frac{3}{4} \sin^{2} 2 x \geq - \frac{3}{4} 1 \geq 1 - \frac{3}{4} \sin^{2} 2 x \geq - \frac{3}{4} + 1 1 \geq 1 - \frac{3}{4} \sin^{2} 2 x \geq \frac{1}{4} \frac{1}{1} \leq \frac{1}{1 - \frac{3}{4} \sin^{2} 2 x} \leq 4 1 \leq \frac{1}{\sin^{6} x + \cos^{6} x} \leq 4 1 \leq y \leq 4 M a x i m u m v a l u e o f y = 4$

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y = \frac{1}{{sin}^{6} x + {cos}^{6} x}

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