<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] I took the long-haul approach for you since it's nice and clear to see. There is a lot of play around with the fact: $\sin^2\theta + \cos^2\theta = 1 $ rearranged into $\sin^2\theta = 1 - \cos^2\theta $ and $\cos^2\theta = 1 - \sin^2\theta $
We can see that: $\cos^4\theta = \cos^2\theta\cos^2\theta = (1-\sin^2\theta)(1-\sin^2\theta) = 1-2\sin^2\theta + \sin^4\theta $
$\sin^2\theta + \cos^4\theta = \cos^2\theta + \sin^4\theta $
$\sin^2\theta + (1-2\sin^2\theta + \sin^4\theta) = \cos^2\theta + \sin^4\theta $
$\sin^2\theta + 1-2\sin^2\theta + \sin^4\theta = \cos^2\theta + \sin^4\theta $
$\sin^4\theta-\sin^2\theta+1= \cos^2\theta + \sin^4\theta $
$\sin^4\theta-(1-\cos^2\theta)+1=\cos^2\theta + \sin^4\theta $
$\sin^4\theta-1+\cos^2\theta+1=\cos^2\theta + \sin^4\theta $
$\sin^4\theta+\cos^2\theta=\cos^2\theta + \sin^4\theta $
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