Szczegółowe wyjaśnienie:
[tex]2(1-\sin^2\alpha)(1-\cos^2\alpha)=1-\sin^4\alpha-\cos^4\alpha\\\\\\L=2\cos^2\alpha\sin^2\alpha=2\sin^2\alpha\cos^2\alpha\\\\P=\sin^2\alpha+\cos^2\alpha-\sin^4\alpha-\cos^4\alpha\\\\=\sin^2\alpha-\sin^4\alpha+\cos^2\alpha-\cos^4\alpha\\\\=\sin^2\alpha(1-\sin^2\alpha)+\cos^2\alpha(1-\cos^2\alpha)\\\\=\sin^2\alpha\cos^2\alpha+\cos^2\alpha\sin^2\alpha=2\sin^2\alpha\cos^2\alpha\\\\L=P\\\\\blacksquare[/tex]
skorzystałem z:
[tex]\sin^2\alpha+\cos^2\alpha=1\to\sin^2\alpha=1-\cos^2\alpha\\\\\sin^2\alpha+\cos^2\alpha=1\to\cos^2\alpha=1-\sin^2\alpha[/tex]
