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?
 A: As $\cos^2 \theta +\sin^2 \theta= 1$ we have  $$\cos^4 \theta -\sin^4 \theta =(\cos^2 \theta -\sin^2 \theta)\color{red}{(\cos^2 \theta +\sin^2 \theta)}= \cos^2 \theta -\sin^2 \theta$$
That is, 
$$\sin^2 \theta +\cos^4 \theta =\cos^2 \theta +\sin^4 \theta$$
A: 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 $
A: Here would be the other points to remember:
$sin^2\theta+cos^2\theta=1$
$x^4-y^4=(x^2-y^2)(x^2+y^2)$
A: Rewrite this as
$$
\sin^2 \theta - \cos^2 \theta = \sin^4 \theta - \cos^4 \theta
$$
and then factor the right-hand side as a difference of two squares.
A: Here's an alternative. I'm not quite sure if this qualifies as a proof, but I think it's an interesting fact:
Consider the function
$$f(\theta) = \sin^2{\theta} - \cos^2{\theta} - \sin^4 \theta + \cos^4 \theta $$
Thus, $f'(\theta) = 4 \sin\theta \cos\theta \, ( 1 - \sin^2\theta - \cos^2 \theta ) = 0$ and $f$ is therefore constant for any $\theta$. We discover this constant is 0 since $f(0) = 0$.
Hope you find this useful/interesting.
A: A "forwards" proof:
Render
$sin^2\theta+\cos^4\theta=sin^2\theta+(1-cos^2\theta)^2=sin^2\theta+(1-2\sin^2\theta+sin^4\theta)$
Regroup the terms on the right as 
$(sin^2\theta+1-2sin^2\theta)+\sin^4\theta$
and put $1-\sin^2\theta=\cos^2\theta$.
A: $$\sin^2\theta+\cos^4\theta=\sin^2\theta+\bigg(\cos^2\theta\bigg)^2=\sin^2\theta+\bigg(1-\sin^2\theta\bigg)^2=\dots$$
A: You can prove this fact by showing that the LHS and RHS agree at the point 0 and that their derivatives are equal everywhere.
Let's use the following as abbreviations for sine and cosine.
$$ y(t) = \sin(t) $$
$$ x(t) = \cos(t) $$
We want to show (Orig).
$$ y^2 + x^4 = x^2 + y^4 \tag{Orig} $$
Negate the goal.
$$ y^2 + x^4 \neq x^2 + y^4 \tag{NG} $$
If we can show that $y(0)^2 + x(0)^4 \neq x(0)^2 + y(0)^4 $ (101) and that the derivatives of both sides are unequal (105), that's equivalent to (NG).
$$ y(0)^2 + x(0)^4 \neq x(0)^2 + y(0)^4 \tag{101} $$
$$ 0 + 1 \neq 1 + 0 \tag{102} $$
$$ 1 \neq 1 \tag{103} $$
$$ \bot \tag{104} $$
And the derivative
$$ (y^2 + x^4)' \neq (x^2 + y^4)' \tag{105} $$
$$ 2yy' + 4x^3x' \neq 2xx' + 4y^3y' \tag{106} $$
Normalize $y' \mapsto x$ and $x' \mapsto -y$ .
$$ 2yx - 4x^3y \neq -2yx + 4y^3x \tag{107} $$
Combine like terms.
$$ 4yx \neq 4y^3x + 4x^3y \tag{108} $$
$$ 4yx \neq 4yx\cdot(y^2 + x^2) \tag{109} $$
$$ 4yx \neq 4yx\cdot 1 \tag{110} $$
$$ 4yx \neq 4yx \tag{111} $$
$$ \bot \tag{112} $$
A: $$ \sin^2\theta + \cos^4\theta = \cos^2\theta + \sin^4\theta$$
$$ \Longleftrightarrow $$
$$ \sin^2\theta - \cos^2\theta = \sin^4\theta - \cos^4\theta$$
$$ \Longleftrightarrow $$
$$ \sin^2\theta - \cos^2\theta = (\underbrace{\sin^2\theta + \cos^2\theta}_{1})(\sin^2\theta - \cos^2\theta) $$
$$ \Longleftrightarrow $$
$$ \sin^2\theta - \cos^2\theta = \sin^2\theta - \cos^2\theta $$
A: Bringing everything in terms of $\cos \theta=c: $
$$LHS=1-c^2+c^4\,;$$
$$RHS= c^2+(1-c^2)^2= c^2 +1-2 c^2 +c^4=1-c^2+c^4\,=LHS\,$$
