Show that \(\sin^4 \theta - \cos^4 \theta \equiv 2 \sin^2 \theta - 1\).
Solution
We start with the expression \(\sin^4 \theta - \cos^4 \theta\).
Using the identity \(a^2 - b^2 = (a-b)(a+b)\), we can rewrite it as:
\((\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta)\).
Since \(\sin^2 \theta + \cos^2 \theta = 1\), the expression simplifies to:
\(\sin^2 \theta - \cos^2 \theta\).
We know that \(\cos^2 \theta = 1 - \sin^2 \theta\), so:
\(\sin^2 \theta - (1 - \sin^2 \theta) = 2\sin^2 \theta - 1\).
Thus, \(\sin^4 \theta - \cos^4 \theta \equiv 2 \sin^2 \theta - 1\).
Log in to record attempts.