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Question Number 177792 by cortano1 last updated on 09/Oct/22

$$\mathrm{Let}{\mathrm{f}}_{\mathrm{k}}\left(\mathrm{x}\right)=\frac{1}{\mathrm{k}}(\sin {}^{\mathrm{k}}\left(\mathrm{x}\right)+\cos {}^{\mathrm{k}}\left(\mathrm{x}\right))$$$$\mathrm{for}\mathrm{k}=1,2,3,....$$$$\mathrm{Prove}\mathrm{that}$$$${\mathrm{f}}_4\left(\mathrm{x}\right)-{\mathrm{f}}_6\left(\mathrm{x}\right)=\frac{1}{12}\mathrm{for}\mathrm{all}\mathrm{real}\mathrm{numbers}\mathrm{x}$$

Answered by mahdipoor last updated on 09/Oct/22

$$get{sin}^{2}x=a\mathrm{\&}{cos}^{2}x=b$$$$\Rightarrow a+b=1\Rightarrow ab={\left(\frac{sin2x}{2}\right)}^2=c$$$$f_4\left(x\right)=\frac{1}{4}(a^2+b^2)=\frac{1}{4}[{(a+b)}^2-2ab]$$$$\frac{1}{4}(1-2c)$$$$f_6=\frac{1}{6}(a^3+b^3)=\frac{1}{6}[{(a+b)}^3-3{ab}^2-3{ba}^2]=$$$$\frac{1}{6}({(a+b)}^3-3ab(a+b))=\frac{1}{6}(1-3c)$$$$f_4-f_6=\frac{1}{4}-\frac{1}{6}=\frac{1}{12}$$

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