Simplify: (sin theta - cos theta)^2 - (Sin theta + cos theta)^2
\[(\sin \theta -\cos \theta)^2\] is same as \[(\sin \theta -\cos \theta )(\sin \theta -\cos \theta)\] Foil!
is the the answer -4sin theta cos theta?
i would like to see the work 2! :)
Using foil I get sin^2 theta - 2sin theta cos theta + cos^2 theta -( sin^2 theta +2 sin theta cos theta + cos^2 theta)
very good :=))
Thanks!
yw! gO_Od job! :)
\((\sin \theta - \cos \theta)^2 - (\sin \theta + \cos \theta)^2\) \(=\sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta - (\sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta)\) \(=\sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta - \sin^2 \theta - 2\sin \theta \cos \theta - \cos^2 \theta\) \(=\color{red}{\sin^2 \theta} - 2\sin \theta \cos \theta + \color{green}{\cos^2 \theta }\color{red}{- \sin^2 \theta} - 2\sin \theta \cos \theta \color{green}{- \cos^2 \theta}\) \(=\cancel{\color{red}{\sin^2 \theta}} - 2\sin \theta \cos \theta + \cancel{\color{green}{\cos^2 \theta }}\color{red}{- \cancel{\sin^2 \theta}} - 2\sin \theta \cos \theta \color{green}{- \cancel{\cos^2 \theta}}\) \(= - 2\sin \theta \cos \theta - 2\sin \theta \cos \theta\) \(= - 4\sin \theta \cos \theta\)
(sin ø - cos ø)^2 - (sin ø + cos ø)^2 = (1 - sin 2ø) - (1 + sin 2ø) = - 2 sin 2ø = -4 sin ø cos ø
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