Question Number 190893 by mnjuly1970 last updated on 13/Apr/23
$$ \\ $$$$\:\:\:\:\:\:\:\:{if}\:,\:\:\frac{{sin}\left({x}\right)−{cos}\left({x}\right)}{{sin}\left({x}\right)+{cos}\left({x}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\Rightarrow\:{find}\:\:{the}\:{value}\:{of} \\ $$$$\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{F}\:=\:{sin}^{\:\mathrm{6}} \left({x}\right)\:+\:{cos}^{\:\mathrm{6}} \left({x}\right)=\:? \\ $$$$ \\ $$
Answered by Frix last updated on 13/Apr/23
$$\Rightarrow\:\mathrm{tan}\:{x}\:=\frac{\mathrm{5}}{\mathrm{3}}\:\Rightarrow\:{F}=\frac{\mathrm{481}}{\mathrm{1156}} \\ $$
Answered by mehdee42 last updated on 13/Apr/23
$${tanx}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$${F}=\left({sin}^{\mathrm{2}} {x}+{cos}^{\mathrm{2}} {x}\right)\left({sin}^{\mathrm{4}} +{cos}^{\mathrm{4}} {x}−{sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x}\right) \\ $$$$=\left({sin}^{\mathrm{2}} {x}+{cos}^{\mathrm{2}} {x}\right)^{\mathrm{2}} −\mathrm{3}{sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x} \\ $$$$=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}{sin}^{\mathrm{2}} \mathrm{2}{x}=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}×\left(\frac{\mathrm{2}{tanx}}{\mathrm{1}+{tan}^{\mathrm{2}} {x}}\right)^{\mathrm{2}} =… \\ $$