Let fk(x)=1ksink⁡x+cosk⁡x where x∈R and k≥1 then f4(x)−f6(x) is equal to .

Solution:

$f_{4} (x) - f_{6} (x) = \frac{1}{4} (\sin^{4} x + \cos^{4} x) - \frac{1}{6} (\cos^{6} x + \sin^{6} x)$

$= \frac{1}{4} \{{(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} x \cos^{2} x\}$ $- \frac{1}{6} \{{(\cos^{2} x + \sin^{2} x)}^{3} - 3 \cos^{2} x \sin^{2} x\}$

$= \frac{1}{4} (1 - 2 \sin^{2} x \cos^{2} x) - \frac{1}{6} (1 - 3 \sin^{2} x \cos^{2} x) = \frac{1}{4} - \frac{1}{6} = \frac{1}{12}$

Let $f_{k} (x) = \frac{1}{k} (\sin^{k} x + \cos^{k} x)$ where $x \in R$ and $k \geq 1$ then $f_{4} (x) - f_{6} (x)$ is equal to .

Solution:

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