Question Number 30856 by Joel578 last updated on 27/Feb/18
Commented by mrW2 last updated on 28/Feb/18
$${Question}\:{is}\:{wrong}.\:{It}\:{should}\:{be}: \\ $$$$\mathrm{sin}^{\mathrm{4}} \:{A}=… \\ $$
Answered by MJS last updated on 27/Feb/18
$$\mathrm{none}\:\mathrm{of}\:\mathrm{these} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos2}{A}=\mathrm{cos}^{\mathrm{2}} {A}−\frac{\mathrm{1}}{\mathrm{2}}={U} \\ $$$$\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cos4}{A}=\frac{\mathrm{1}}{\mathrm{8}}−\mathrm{sin}^{\mathrm{2}} {A}\mathrm{cos}^{\mathrm{2}} {A}={V} \\ $$$${U}+{V}=\mathrm{cos}^{\mathrm{4}} {A}−\frac{\mathrm{3}}{\mathrm{8}} \\ $$$${U}−{V}=\frac{\mathrm{3}}{\mathrm{8}}−\mathrm{sin}^{\mathrm{4}} {A} \\ $$