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How do you simplify sin^4-cos^4=2sin^2-1?

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\bf \textit{Double Angle Identities} \\\\ cos(2\theta)= \begin{cases} cos^2(\theta)-sin^2(\theta)\\ 1-2sin^2(\theta)\\ 2cos^2(\theta)-1 \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}


\bf \begin{array}{llll} sin^4(\theta )-cos^4(\theta )&=&2sin^2(\theta )-1\\[1em] [sin(\theta )]^4-[cos(\theta )]^4&=&-[1-2sin^2(\theta )]\\[1em] \stackrel{\textit{difference of squares}}{[[sin(\theta )]^2]^2-[[cos(\theta )]^2]^2}&=&-[cos^2(\theta )-sin^2(\theta )]\\[1em] [sin^2(\theta )-cos^2(\theta )][sin^2(\theta )+cos^2(\theta )]&=&sin^2(\theta )-cos^2(\theta )\\[1em] [sin^2(\theta )-cos^2(\theta )][1]&=\\\\ sin^2(\theta )-cos^2(\theta ) &=\end{array}

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