Space span of $f=\cos^2(x)$ and $g=\sin^2(x)$? I know how to determine if vectors/matrices/polynomials span a vector space, even though I don't really understand the concept. Now I'm facing the question of which ones of the next options are within the span of $f=\cos^2(x)$ and $g=\sin^2(x)$:
1) $\cos(2x)$
2) $3+x^2$
3) $1$
4) $\sin(x)$  
My guess is that only 3 and 4 are in the space spanned by f and g, but I can't really explain it or prove it. With vectors/matrices/polynomials I use the matrix method (I don't know if it has a name), but I can't use that one with these. What do I need to know to understand how this concept works with trigonometric functions? Sorry if it's a dumb question.
 A: We have $\cos 2x=\cos^2 x-\sin^2 x$, so $\cos 2x$ is in the space spanned by $\cos^2 x$ and $\sin^2 x$. 
The function $\sin x$ is not in the space spanned. There are many ways to see this. For example, $\sin x$ is an odd function, that is, $\sin(-x)=-\sin x$. But $\cos^2x$ and $\sin^2 x$ are even functions, and therefore so is any linear combination of them. Another way of seeing it is that $\sin x$ has smallest period $2\pi$, while $\cos^2 x$ and $\sin^2 x$ have smallest period $\pi$. 
Or else we can suppose that $\sin x=a\cos^2 x+b\sin^2 x$ for some real numbers $a$ and $b$, and then use special values of the trig functions to derive a contradiction. As a start, let $x=0$. That tells us that $a=0$. 
You are right in deciding $3+x^2$ is not in the span. Any function in the span is periodic, and $3+x^2$ isn't.
A: Clearly $(1)$ and $(3)$ are in the linear span of $f$ and $g$, since


*

*$\cos 2x=\cos^2 x-\sin^2 x$

*$\cos^2x+\sin^2x=1$


For the other options, 
Checking for $\sin x$,
if $a\cos^2 x+b\sin^2 x=\sin x$
then $$a(1-sin^2 x)+b\sin^2 x=\sin x$$
$$\implies (a-b)\sin^2 x + \sin x - a=0$$
Does there exists a solution, for some $a,b\in \mathbb R$?
And for $1+x^2$,
Let $$a \sin ^2 x+b\cos^2 x = 3+x^2$$
for $x=0,$ we get $ b=3$ and for $x=\frac\pi 2 $ we get $a= 3+(\frac \pi 2) ^2 $
And for $x=\frac \pi 4$, we $\frac a 2 + \frac b 2 =1\frac 1 2 . i.e., a+b=3$, which gives so many solutions.
So, $1+x^2\notin  span 
\{f,g\}$.
A: Hint: write $$a\cos^2(x)+b\sin^2(x)=blah$$
Plug in values of $x$ like $0$, $\pi/2$ to solve for $a$, $b$. Then see if the equation is true. 
