hello
what's the dimension of the subspace $\displaystyle F$ of $\displaystyle \mathfrak{F}(\mathbb{R},\mathbb{R}) $ spanned by $\displaystyle f_1(x)=\sin^2(x),f_2(x)=\cos^2(x),f_3(x)=\sin(2x)$ and $\displaystyle f_4(x)=\cos(2x)$
thanks.
you should explain very clearly what your vector space is. just using a non-standard and weird notation such as $\displaystyle \mathfrak{F}(\mathbb{R},\mathbb{R}) $ is not good enough. i guess your vector space is the $\displaystyle \mathbb{R}$-vector space of
all real-valued functions defined on $\displaystyle \mathbb{R}$. then $\displaystyle \dim_{\mathbb{R}}F=2$ and $\displaystyle \{\sin^2 x, \sin (2x) \}$ is a basis for $\displaystyle F.$
Edit: i meant $\displaystyle \dim_{\mathbb{R}} F=3$ and $\displaystyle \{1,\sin^2x, \sin(2x) \}$ is a basis for $\displaystyle F.$ sorry for the mistake.
I don't think this is right since assume there exist $\displaystyle a,b \in \mathbb{R}$ such that $\displaystyle \cos ^2 (x)= a\sin (2x) + b\sin ^2 (x)=2a \sin (x) \cos (x) + b\sin ^2 (x)$ then evaluation in $\displaystyle 0$ gives $\displaystyle 1=0$. Thus $\displaystyle A= \{ \sin ^2 (x), \cos ^2 (x) , \sin (2x) \}$ is l.i. and since $\displaystyle \cos ^2 (x) - \sin ^2 (x) = \cos (2x)$ we get that $\displaystyle A$ is a basis for $\displaystyle F$.
Assume $\displaystyle \sin ^2(x) = a \sin (2x)$ for some $\displaystyle a\in \mathbb{R}$ then evaluation in $\displaystyle \frac{\pi }{2}$ will give $\displaystyle 1=0$ so these two functions are l.i. Then by my previous post we get that $\displaystyle \cos ^2 (x) \notin span \{ \sin ^2 (x), \sin (2x) \}$ so it must be that these three are l.i. but again by my previous post $\displaystyle \cos (2x) \in span(A)$ so $\displaystyle A$ is a basis for $\displaystyle F$.