Question Number 83036 by otchereabdullai@gmail.com last updated on 27/Feb/20
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{cos}^{\mathrm{4}} \theta−\mathrm{sin}^{\mathrm{4}} \theta=\mathrm{cos}^{\mathrm{2}} \theta−\mathrm{sin}^{\mathrm{2}} \theta \\ $$
Commented by Tony Lin last updated on 27/Feb/20
$${cos}^{\mathrm{4}} \theta−{sin}^{\mathrm{4}} \theta \\ $$$$=\left({cos}^{\mathrm{2}} \theta−{sin}^{\mathrm{2}} \theta\right)^{\mathrm{2}} +\mathrm{2}{sin}^{\mathrm{2}} \theta{cos}^{\mathrm{2}} \theta \\ $$$$=\frac{{cos}\mathrm{4}\theta+{sin}\mathrm{4}\theta+\mathrm{1}}{\mathrm{2}}\neq{cos}\mathrm{2}\theta \\ $$$$ \\ $$
Commented by MJS last updated on 27/Feb/20
$$\mathrm{cos}\:\theta\:={c}\wedge\mathrm{sin}\:\theta\:={s} \\ $$$${c}^{\mathrm{4}} −{s}^{\mathrm{4}} ={c}^{\mathrm{2}} −{s}^{\mathrm{2}} \\ $$$$\left({c}^{\mathrm{2}} −{s}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} +{s}^{\mathrm{2}} \right)={c}^{\mathrm{2}} −{s}^{\mathrm{2}} \\ $$$$\mathrm{but}\:{c}^{\mathrm{2}} +{s}^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{true} \\ $$
Commented by otchereabdullai@gmail.com last updated on 27/Feb/20
$$\mathrm{thanks}\:\mathrm{prof}\:\mathrm{mjs} \\ $$