Discussion Forum

 If   $f_{k}(x)=1/k(\sin^{k}x+\cos^{k}x)$ , where x ε  R and  $k\geq 1$   , then  $f_{4}(x)-f_{6}(x)$   is equal to 

Answer:

Explanation:

Given     $f_{k}(x)=1/k(\sin^{k}x+\cos^{k}x)$ , where x ε  R and k>1

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

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

   =  $\frac{1}{4}-\frac{1}{6}=\frac{1}{12}$