Question Number 196639 by bbbbbbbb last updated on 29/Aug/23
$$\mathrm{if}\:\mathrm{sinx}=\frac{\mathrm{2}}{\mathrm{3}}\:\:\boldsymbol{\mathrm{then}}\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{sin}}^{\mathrm{6}} \boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{cos}}^{\mathrm{6}} \boldsymbol{\mathrm{x}}??? \\ $$
Answered by deleteduser1 last updated on 28/Aug/23
$$=\left({sin}^{\mathrm{2}} {x}+{cos}^{\mathrm{2}} {x}\right)^{\mathrm{3}} −\mathrm{3}{sin}^{\mathrm{2}} {cos}^{\mathrm{2}} {x}\left({sin}^{\mathrm{2}} {x}+{cos}^{\mathrm{2}} {x}\right) \\ $$$$=\mathrm{1}−\mathrm{3}{sin}^{\mathrm{2}} \left({x}\right)\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)\left(\mathrm{1}\right)=\mathrm{1}−\mathrm{3}\left(\frac{\mathrm{4}}{\mathrm{9}}\right)\left(\frac{\mathrm{5}}{\mathrm{9}}\right)=\frac{\mathrm{7}}{\mathrm{27}} \\ $$
Answered by Frix last updated on 28/Aug/23
$${s}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${c}=\pm\sqrt{\mathrm{1}−{s}^{\mathrm{2}} }=\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{3}} \\ $$$${s}^{\mathrm{6}} +{c}^{\mathrm{6}} =\frac{\mathrm{7}}{\mathrm{27}} \\ $$$$ \\ $$$${s}^{\mathrm{6}} +{c}^{\mathrm{6}} ={s}^{\mathrm{6}} +\left(\mathrm{1}−{s}^{\mathrm{2}} \right)^{\mathrm{3}} =\mathrm{3}{s}^{\mathrm{4}} −\mathrm{3}{s}^{\mathrm{2}} +\mathrm{1}=\frac{\mathrm{7}}{\mathrm{27}} \\ $$