Let f_k(x) = frac{1}{k}(sin^k x + cos^k x),wh | vedclass

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(B) Given $f_k(x) = \frac{1}{k}(\sin^k x + \cos^k x)$.We need to find $f_4(x) - f_6(x)$.$f_4(x) = \frac{1}{4}(\sin^4 x + \cos^4 x) = \frac{1}{4}((\sin

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$\frac{1}{12}$

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(B) Given $f_k(x) = \frac{1}{k}(\sin^k x + \cos^k x)$.We need to find $f_4(x) - f_6(x)$.$f_4(x) = \frac{1}{4}(\sin^4 x + \cos^4 x) = \frac{1}{4}((\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x) = \frac{1}{4}(1 - 2\sin^2 x \cos^2 x)$.$f_6(x) = \frac{1}{6}(\sin^6 x + \cos^6 x) = \frac{1}{6}((\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x)) = \frac{1}{6}(1 - 3\sin^2 x \cos^2 x)$.Now,$f_4(x) - f_6(x) = \frac{1}{4}(1 - 2\sin^2 x \cos^2 x) - \frac{1}{6}(1 - 3\sin^2 x \cos^2 x)$.$= (\frac{1}{4} - \frac{1}{6}) - (\frac{2}{4} - \frac{3}{6})\sin^2 x \cos^2 x$.$= (\frac{3-2}{12}) - (\frac{1}{2} - \frac{1}{2})\sin^2 x \cos^2 x$.$= \frac{1}{12} - 0 = \frac{1}{12}$.

Let $f_k(x) = \frac{1}{k}(\sin^k x + \cos^k x)$,where $x \in R$ and $k \ge 1$. Then $f_4(x) - f_6(x)$ is equal to

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