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

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(B) Given $f_k(x) = \frac{1}{k}(\cos^k x + \sin^k x)$.For $k=4$, $f_4(x) = \frac{1}{4}(\cos^4 x + \sin^4 x)$.Using $\cos^4 x + \sin^4 x = (\cos^2 x +

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$-\frac{1}{12}$

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(B) Given $f_k(x) = \frac{1}{k}(\cos^k x + \sin^k x)$.For $k=4$, $f_4(x) = \frac{1}{4}(\cos^4 x + \sin^4 x)$.Using $\cos^4 x + \sin^4 x = (\cos^2 x + \sin^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}\sin^2(2x)$, we get $f_4(x) = \frac{1}{4}(1 - \frac{1}{2}\sin^2(2x))$.For $k=6$, $f_6(x) = \frac{1}{6}(\cos^6 x + \sin^6 x)$.Using $\cos^6 x + \sin^6 x = (\cos^2 x + \sin^2 x)(\cos^4 x - \cos^2 x \sin^2 x + \sin^4 x) = 1 - 3\sin^2 x \cos^2 x = 1 - \frac{3}{4}\sin^2(2x)$, we get $f_6(x) = \frac{1}{6}(1 - \frac{3}{4}\sin^2(2x))$.Now, $f_6(x) - f_4(x) = \frac{1}{6} - \frac{1}{8}\sin^2(2x) - (\frac{1}{4} - \frac{1}{8}\sin^2(2x)) = \frac{1}{6} - \frac{1}{4} = \frac{2-3}{12} = -\frac{1}{12}$.

Let $f_k(x) = \frac{1}{k}(\cos^k x + \sin^k x)$ where $k \in \mathbb{N}$. Find the value of $f_6(x) - f_4(x)$.

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