Question

Aratati ca 2 (sin^6 a +cos^6 a ) -3 (sin^4 a +cos^4 a ) =-1 , oricare a € R . Mentionez ca puterile sunt reprezentate de 6 si 4 , ci nu de 6a sau 4a . Multumesc

Answer 1

$2(\sin^{6}{a}+\cos^{6}{a})-2(\sin^{4}{a}+\cos^{4}{a})-(\sin^{4}{a}+\cos^{4}{a})=2\sin^{4}{a}(\sin^{2}{a}-1)+2\cos^{4}{a}(\cos^{2}{a}-1)-(\sin^{4}{a}+\cos^{4}{a})=2\sin^{4}{a}(\sin^{2}{a}-\sin^{2}{a}-\cos^{2}{a})+2\cos^{4}{a}(\cos^{2}{a}-\sin^{2}{a}-\cos^{2}{a})-(\sin^{4}{a}+\cos^{4}{a})=-2\sin^{4}{a}\cos^{2}{a}-2\cos^{4}{a}\sin^{2}{a}-(\sin^{4}{a}+\cos^{4}{a})=-2\sin^{2}{a}\cos^{2}{a}(\sin^{2}{a}+\cos^{2}{a})-(\sin^{4}{a}+\cos^{4}{a})=-(\sin^{4}{a}+2\sin^{2}{a}\cos^{2}+\cos^{4}{a})=-(\sin^{2}{a}+\cos^{2}{a})^{2}=-1^{2}=-1$