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Let $f_{k}{x}=\frac{1}{k}\left ( sin^{k}+cos^{k}x \right )$ for k = 1,2,3,..... Then for all x $\varepsilon$ R , the value of $f_{4}(x)-f_{6}(x)$ is equal to :
Option 1)

$\frac{1}{4}$
Option 2)

$\frac{-1}{12}$
Option 3)

$\frac{5}{12}$

Option 4)

$\frac{1}{12}$

Answers (1)

best_answer

Trigonometric Identities -

$\sin ^{2}\Theta + \cos ^{2}\Theta = 1$

$1 + \tan ^{2}\Theta = \sec ^{2}\Theta$

$1 + \cot ^{2}\Theta = cosec ^{2}\Theta$

- wherein

They are true for all real values of $\Theta$

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

$=\frac{1}{4}\left [ \left ( \sin ^{2} +\cos ^{2}x\right )^{2} -2\sin ^{2}x\cos ^{2}x\right ]$

$=\frac{1}{4}-\frac{1}{2}\sin ^{2}x\cos ^{2}x$

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

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

$=\frac{1}{6}\left [ 1-3\sin ^{2}x\cos ^{2}x\left ( \sin ^{2}x+\cos ^{2}x \right ) \right ]$

$=\frac{1}{6}-\frac{1}{2}\sin ^{2}x\cos ^{2}x$

$f_{4}(x)-f_{6}(x)=\frac{1}{4}-\frac{1}{6}=\frac{1}{12}$

M

Option 1)

$\frac{1}{4}$

Option 2)

$\frac{-1}{12}$

Option 3)

$\frac{5}{12}$

Option 4)

$\frac{1}{12}$

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