Question Number 104894 by bramlex last updated on 24/Jul/20
$${prove}\:{that}\:\mathrm{cos}\:^{\mathrm{6}} {a}\:−\mathrm{sin}\:^{\mathrm{6}} {a}\:=\: \\ $$$$\mathrm{cos}\:\mathrm{2}{a}\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{a}\right)\: \\ $$
Answered by john santu last updated on 24/Jul/20
$$\left(\mathrm{1}\right)\mathrm{cos}\:^{\mathrm{6}} {a}−\mathrm{sin}\:^{\mathrm{6}} {a}\:=\:\left(\mathrm{cos}\:^{\mathrm{2}} {a}−\mathrm{sin}\:^{\mathrm{2}} {a}\right) \\ $$$$\left(\mathrm{cos}\:^{\mathrm{4}} {a}+\mathrm{sin}\:^{\mathrm{2}} {a}\:\mathrm{cos}\:^{\mathrm{2}} {a}+\mathrm{sin}\:^{\mathrm{4}} {a}\right) \\ $$$$=\left(\mathrm{cos}\:^{\mathrm{2}} {a}−\mathrm{sin}\:^{\mathrm{2}} {a}\right)\left\{\left(\mathrm{cos}\:^{\mathrm{2}} {a}+\mathrm{sin}\:^{\mathrm{2}} {a}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2sin}\:{a}\:\mathrm{cos}\:{a}\right)^{\mathrm{2}} \right\} \\ $$$$=\:\mathrm{cos}\:\mathrm{2}{a}\:\left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{2}{a}\:\right\}\:\left({JS}\:\circledast\right) \\ $$