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Mathematics
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Establish the following trig identity
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\[\sin^4\theta - \cos^4\theta = 1 - 2\cos^2\theta\]
Equations: \[\tan \theta = \frac{ \sin \theta }{ \cos \theta }\] \[\cot \theta = \frac{ \cos \theta }{ \sin \theta }\] \[\sec \theta = \frac{ 1 }{ \cos \theta}\] \[\csc \theta = \frac{ 1 }{ \sin \theta}\] \[\cot \theta = \frac{ 1 }{ \tan \theta}\]
\[\sin^2 \theta + \cos^2 \theta = 1\]
hint: \[A^2-B^2 = (A+B)(A-B)\]
and \[(sin^2(\theta))^2 = sin^4(\theta)\]
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Thank you. I think I've got it. \[= ( \sin^2 \theta + \cos^2 \theta ) (\sin^2 \theta - \cos^2 \theta)\] Then just simplify to get sin^2 theta = 1 - cos^2 theta
yep. :)
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