[tex]\sin^4x-\cos^4x=1-2\cos^2x\\\\
\sin^4x-\cos^4x=1-(\cos^2x+\cos^2x)\\\\
\sin^4x-\cos^4x=1-\cos^2x-\cos^2x\\\\
\sin^4x-\cos^4x=\sin^2x-\cos^2x\\\\
(\sin^2x+\cos^2x)(\sin^2x-\cos^2x)=\sin^2x-\cos^2x[/tex]
Dividindo-se toda equação por [tex]\sin^2x-\cos^2x[/tex], temos:
[tex]\sin^2x+\cos^2=1\Longrightarrow OK!\;\blacksquare[/tex]
