Question

the maximum value of cos^6 X +sin^6 X is
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Answer 1

Answer:

Maximum value of cos⁶(x) + sin⁶(x) is 1.

Step-by-step explanation:

$\sin {}^{6} (x) + \cos {}^{6} (x)$

$= {( \sin {}^{2} (x) )}^{3} + {( \cos {}^{2} (x) )}^{3}$

Using the Identity

${a}^{3} + {b}^{3} = {(a + b)}^{3} - 3ab(a + b)$

we get

$= { (\sin {}^{2} (x) + \cos {}^{2} (x) ) }^{3} - 3 \sin {}^{2} (x) \cos {}^{2} (x) ((\sin {}^{2} (x) + \cos {}^{2} (x))$

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

$= 1 - \frac{3}{4} {(2 \sin(x) \cos(x)) }^{2}$

$= 1 - \frac{3}{4} {( \sin(2x) )}^{2}$

$= 1 - \frac{3}{4} \sin {}^{2} (2x)$

Since

$- 1 \leqslant \sin(2x) \leqslant 1$

$or \: 0 \leqslant \sin {}^{2} (2x) \leqslant 1$

$or \: 0 \leqslant \frac{3}{4 } \sin {}^{2} (2x) \leqslant \frac{3}{4}$

$or \: 0 \geqslant - \frac{3}{4} \sin {}^{2} (2x) \geqslant - \frac{3}{4}$

$or \: 1 \geqslant 1 - \frac{3}{4} \sin {}^{2} (2x) \geqslant 1 - \frac{3}{4}$

$or \: \frac{1}{4} \leqslant 1 - \frac{3}{4} \sin {}^{2} (2x) \leqslant 1$

$\: \frac{1}{4} \leqslant \ \cos {}^{6} (x) + \sin {}^{6} (x) \leqslant 1$

Therefore,Maximum and minimum value of cos⁶(x) + sin⁶(x) are 1 and 1/4 respectively.

Answer 2

Answer:

the range of sin⁶ x +. cos⁶ x is