<s>[INST] Prove $\sin^2 \theta +\cos^4 \theta =\cos^2 \theta +\sin^4 \theta $ Prove $$\sin^2(\theta)+\cos^4(\theta)=\cos^2(\theta)+\sin^4(\theta)$$
I only know how to solve using factoring and the basic trig identities, I do not know reduction or anything of the sort, please prove using the basic trigonometric identities and factoring.
After some help I found that you move the identity around, so:
$\sin^2(\theta)-\cos^2(\theta)=\sin^4(\theta)-\cos^4(\theta)$
Then,
$\sin^2(\theta)-\cos^2(\theta)=(\sin^2(\theta)+\cos^2(\theta))(\sin^2(\theta)-\cos^2(\theta))$
the positive sum of squares defaults to 1 and then the right side equals the left, but how does that prove the original identity?
 [/INST] $$ \sin^2\theta + \cos^4\theta = \cos^2\theta + \sin^4\theta$$
$$ \Longleftrightarrow $$
$$ \sin^2\theta - \cos^2\theta = \sin^4\theta - \cos^4\theta$$
$$ \Longleftrightarrow $$
$$ \sin^2\theta - \cos^2\theta = (\underbrace{\sin^2\theta + \cos^2\theta}_{1})(\sin^2\theta - \cos^2\theta) $$
$$ \Longleftrightarrow $$
$$ \sin^2\theta - \cos^2\theta = \sin^2\theta - \cos^2\theta $$
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