$\sin^2(x), \cos^2(x),$ and $\sin^4(x)$ and linear dependence Since
$$
\sin^{4}x=\sin^{2}x\sin^{2}x = \sin^{2}x(1-\cos^{2}x) = \sin^{2}x-\sin^{2}x\cos^{2}x
$$
does this mean that 
$$
\sin^{2}x,\cos^{2}x, \text{ and } \sin^{4}x
$$
are linearly dependent?

No. Linear dependence would require real numbers $a,b,c$, at least one of which is nonzero, such that $a\sin^2(x)+b\cos^2(x)+c\sin^4(x)=0$. No such real numbers exist. However, the three functions are algebraically dependent, as you have shown.
From the theory of Fourier series, we know that sufficiently well-behaved functions on $[0,2\pi]$ can be uniquely expressed as a sum of the form
$$\sum_{k=-\infty}^{\infty}{e^{kix}}$$
By Euler's formula we have that
$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$
and
$$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$
So
$$\sin^4(x)=\frac{(e^{ix}-e^{-ix})^4}{16}=\frac{1}{16}(e^{4ix}-4e^{2ix}+6+4e^{-2ix}+e^{-4ix})$$
$$\sin^2(x)=\frac{(e^{ix}-e^{-ix})^2}{-4}=-\frac{1}{4}(e^{2ix}-2+e^{-2ix})$$
$$\cos^2(x)=\frac{(e^{ix}+e^{-ix})^2}{4}=\frac{1}{4}(e^{2ix}+2+e^{-2ix})$$
Since the Fourier series of $\sin^4(x)$ has a term of the form $e^{4ix}$ and neither of $\sin^2(x)$ nor $\cos^2(x)$ has such a term, there is no way to express $\sin^4(x)$ as a linear combination of $\sin^2(x)$ and $\cos^2(x)$.
We have left to check whether there exist $a,b$, at least one of which is nonzero, such that $a\cos^2(x)+b\sin^2(x)=0$. In order for that to be the case we would have to have $a=-b$ in order for the terms $e^{2ix}$ and $e^{-2ix}$ to cancel. However, if $a=b\neq 0$ then we have that
$$a\cos^2(x)+a\sin^2(x)=a(\cos^2(x)+\sin^2(x))=a\cdot (1)=a\neq 0$$
so the functions are linearly independent.
