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Question Number 177792 by cortano1 last updated on 09/Oct/22
  Let f_k (x)= (1/k)(sin^k (x)+cos^k (x))    for k=1,2,3,... .     Prove that        f_4 (x)−f_6 (x)= (1/(12)) for all real numbers x
$$\:\:\mathrm{Let}\:\mathrm{f}_{\mathrm{k}} \left(\mathrm{x}\right)=\:\frac{\mathrm{1}}{\mathrm{k}}\left(\mathrm{sin}\:^{\mathrm{k}} \left(\mathrm{x}\right)+\mathrm{cos}\:^{\mathrm{k}} \left(\mathrm{x}\right)\right) \\ $$$$\:\:\mathrm{for}\:\mathrm{k}=\mathrm{1},\mathrm{2},\mathrm{3},…\:.\: \\ $$$$\:\:\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\:\:\:\:\:\mathrm{f}_{\mathrm{4}} \left(\mathrm{x}\right)−\mathrm{f}_{\mathrm{6}} \left(\mathrm{x}\right)=\:\frac{\mathrm{1}}{\mathrm{12}}\:\mathrm{for}\:\mathrm{all}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{x} \\ $$
Answered by mahdipoor last updated on 09/Oct/22
get  sin^2 x=a    &   cos^2 x=b  ⇒a+b=1     ⇒ab=(((sin2x)/2))^2 =c  f_4 (x)=(1/4)(a^2 +b^2 )=(1/4)[(a+b)^2 −2ab]  (1/4)(1−2c)  f_6 =(1/6)(a^3 +b^3 )=(1/6)[(a+b)^3 −3ab^2 −3ba^2 ]=  (1/6)((a+b)^3 −3ab(a+b))=(1/6)(1−3c)  f_4 −f_6 =(1/4)−(1/6)=(1/(12))
$${get}\:\:{sin}^{\mathrm{2}} {x}={a}\:\:\:\:\&\:\:\:{cos}^{\mathrm{2}} {x}={b} \\ $$$$\Rightarrow{a}+{b}=\mathrm{1}\:\:\:\:\:\Rightarrow{ab}=\left(\frac{{sin}\mathrm{2}{x}}{\mathrm{2}}\right)^{\mathrm{2}} ={c} \\ $$$${f}_{\mathrm{4}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)=\frac{\mathrm{1}}{\mathrm{4}}\left[\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}{ab}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}−\mathrm{2}{c}\right) \\ $$$${f}_{\mathrm{6}} =\frac{\mathrm{1}}{\mathrm{6}}\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)=\frac{\mathrm{1}}{\mathrm{6}}\left[\left({a}+{b}\right)^{\mathrm{3}} −\mathrm{3}{ab}^{\mathrm{2}} −\mathrm{3}{ba}^{\mathrm{2}} \right]= \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}\left(\left({a}+{b}\right)^{\mathrm{3}} −\mathrm{3}{ab}\left({a}+{b}\right)\right)=\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{1}−\mathrm{3}{c}\right) \\ $$$${f}_{\mathrm{4}} −{f}_{\mathrm{6}} =\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{12}} \\ $$

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