Math, asked by poojareddy5588, 9 months ago

let FK(x) =1/k(sin to the power of k x+cos to the power of k x) x belongs to R then F4(x) -F6(x)

Answers

Answered by pulakmath007

SOLUTION

GIVEN

$\displaystyle \sf{ F_k(x) = \frac{1}{k} \: ( { \sin}^{k}x + { \cos}^{k} x) }$

TO DETERMINE

The value of

$\displaystyle \sf{ F_4(x) - F_6(x) }$

EVALUATION

Here it is given that

$\displaystyle \sf{ F_k(x) = \frac{1}{k} \: ( { \sin}^{k}x + { \cos}^{k} x) }$

Now

$\displaystyle \sf{ F_4(x) = \frac{1}{4} \: ( { \sin}^{4}x + { \cos}^{4} x) }$

$\displaystyle \sf{ \implies \: F_4(x) = \frac{1}{4} \bigg[{( { \sin}^{2}x + { \cos}^{2} x)}^{2} - 2 { \sin}^{2}x { \cos}^{2} x\bigg] }$

$\displaystyle \sf{ \implies \: F_4(x) = \frac{1}{4} \bigg[1 - 2 { \sin}^{2}x { \cos}^{2} x\bigg] }$

$\displaystyle \sf{ \implies \: F_4(x) = \frac{1}{4} - \frac{1}{2} { \sin}^{2}x { \cos}^{2} x}$

Now

$\displaystyle \sf{ F_6(x) = \frac{1}{6} \: ( { \sin}^{6}x + { \cos}^{6} x) }$

$\displaystyle \sf{ \implies \: F_6(x) = \frac{1}{6} \bigg[{( { \sin}^{2}x + { \cos}^{2} x)}^{3} - 3 { \sin}^{2}x { \cos}^{2} x( { \sin}^{2}x + { \cos}^{2} x)\bigg] }$

$\displaystyle \sf{ \implies \: F_6(x) = \frac{1}{6} \bigg[1 - 3 { \sin}^{2}x { \cos}^{2} x\bigg] }$

$\displaystyle \sf{ \implies \: F_6(x) = \frac{1}{6} - \frac{1}{2} { \sin}^{2}x { \cos}^{2} x}$

Thus

$\displaystyle \sf{ F_4(x) - F_6(x) }$

$\displaystyle \sf{ = \bigg[ \frac{1}{4} - \frac{1}{2} { \sin}^{2}x { \cos}^{2} x\bigg] -\bigg[ \frac{1}{6} - \frac{1}{2} { \sin}^{2}x { \cos}^{2} x\bigg] }$