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  4. If fn(x) = (1/n) ( cosn x + sinn x) , for n = 1,2,3....., then f4(x) - f6 (x) is equal to

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Q. If $f_n(x) = \frac{1}{n} (\cos^{n} x + \sin^{n} x)$ , for $ n = 1,2,3.....,$ then $f_4(x) - f_6 (x)$ is equal to

UPSEEUPSEE 2018

Solution:

Given, $f_{n}\left(x\right)=\frac{1}{n}\left(\cos^{n}\,x + \sin^{n}\,x\right)$,
where $n = 1$, $2$, $3$, ...
Now, $f_{4}\left(x\right)-f_{6}\left(x\right)=\frac{1}{4}\left(cos^{4}\,x + \sin^{4}\,x\right)$
$-\frac{1}{6}\left(cos^{6}\,x + \sin^{6}\,x\right)$
$=\frac{1}{4}\left[\left(\cos^{2}\,x + \sin^{2}\,x\right)^{2}-2\cos^{2}\,x\, \sin^{2}\,x\right]$
$-\frac{1}{6}\left[\left(\cos^{2}\,x\right)^{3}+\left(\sin^{2}\,x\right)^{3}\right]$
$=\frac{1}{4}\left[1-2\sin^{2}\cos^{2}\,x\right]-\frac{1}{6}\left[(\cos^{2}\,x + \sin^{2}\,x\right)^{3}$
$− 3\sin^2 \,x \,\cos^2\, x(\sin^2\, x + \cos^2 \,x)]$
$=\frac{1}{4}\left[1-2\sin^{2}\,x\,\cos^{2}\,x\right]-\frac{1}{6}\left[1-3\sin^{2}\,x\,\cos^{2}\,x\right]$
$=\frac{1}{4}-\frac{1}{2}\sin^{2}\,x\,\cos^{2}\,x-\frac{1}{6}+\frac{1}{2}\sin^{2}\,x\,\cos^{2}\,x$
$=\frac{1}{4}-\frac{1}{6}=\frac{1}{12}$