If f_n(x) = frac{1}{2n} [sin^{2n} x + cos^{2n | vedclass

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(D) Given $f_n(x) = \frac{1}{2n} [\sin^{2n} x + \cos^{2n} x]$.We need to evaluate $f_1(x) + f_2(x) - f_3(x)$.$f_1(x) = \frac{1}{2} [\sin^2 x + \cos^2

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$\frac{7}{12}$

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(D) Given $f_n(x) = \frac{1}{2n} [\sin^{2n} x + \cos^{2n} x]$.We need to evaluate $f_1(x) + f_2(x) - f_3(x)$.$f_1(x) = \frac{1}{2} [\sin^2 x + \cos^2 x] = \frac{1}{2}(1) = \frac{1}{2}$.$f_2(x) = \frac{1}{4} [\sin^4 x + \cos^4 x] = \frac{1}{4} [(\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x] = \frac{1}{4} [1 - 2 \sin^2 x \cos^2 x] = \frac{1}{4} - \frac{1}{2} \sin^2 x \cos^2 x$.$f_3(x) = \frac{1}{6} [\sin^6 x + \cos^6 x] = \frac{1}{6} [(\sin^2 x + \cos^2 x)(\sin^4 x + \cos^4 x - \sin^2 x \cos^2 x)] = \frac{1}{6} [1 - 3 \sin^2 x \cos^2 x]$.Now,$f_1(x) + f_2(x) - f_3(x) = \frac{1}{2} + (\frac{1}{4} - \frac{1}{2} \sin^2 x \cos^2 x) - (\frac{1}{6} - \frac{1}{2} \sin^2 x \cos^2 x)$.$= \frac{1}{2} + \frac{1}{4} - \frac{1}{6} - \frac{1}{2} \sin^2 x \cos^2 x + \frac{1}{2} \sin^2 x \cos^2 x$.$= \frac{6 + 3 - 2}{12} = \frac{7}{12}$.

If $f_n(x) = \frac{1}{2n} [\sin^{2n} x + \cos^{2n} x]$,then $f_1(x) + f_2(x) - f_3(x) =$

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