Let $W$ be the subspace of all real-valued functions spanned by $\{ \cos^2(t), \sin^2(t), \cos(2t) \} $. Find a basis for W. Find the dimension of W. Let $W$ be the subspace of all real-valued functions spanned by $\{ \cos^2(t), \sin^2(t), \cos(2t) \} $. Find a basis for W. Find the dimension of W.
My attempt:
Let $f$ $\in$ $W$. Since $W$ is spanned by the set $\{ \cos^2(t), \sin^2(t), \cos(2t) \}$  therefore any vector of $W$ can be written as a linear combination of these three. So for some $a, b, c \in \mathbb{R}$, one have
$f(t)=a\cos^2(t)+b\sin^2(t)+c\cos(2t)$
Update:
By the trigonometric identity, $\cos(2t)=\cos^2(t)-\sin^2(t)$,
$f(t)=a\cos^2(t)+b\sin^2(t)+c\cos(2t)$
$f(t)=a\cos^2(t)+c\cos^2(t)+b\sin^2(t)-c\sin^2(t)$
$f(t)=(a+c)\cos^2(t)+(b-c)\sin^2(t)$
$f(t)=b_1\cos^2(t)+b_2\sin^2(t)$
where $b_1$ and $b_2 \in \mathbb{R}. $
Due to the fact that $a\cos^2(t)+b\sin^2(t)+c\cos(2t)$ can be written as $b_1\cos^2(t)+b_2\sin^2(t)$ for some $b_1$ and $b_2$, $\{ \cos^2(t), \sin^2(t) \} $ is a spanning set.
Now, is it linearly independent? If $c_1\cos^2(t)+c_2\sin^2(t)=0,$ can there be  $c_1=c_2=0$?
Let $c_1$ and $c_2 \in \mathbb{R}$, $c_1=c_2$, and $c_1, c_2$ $>0$
If $c_1=c_2=2$
$\cos^2(t)+\sin^2(t)=2$
Since $c_1\cos^2(t)+c_2\sin^2(t)=0$ is only true when $c_1=c_2=0$ then it is linearly independent.
Because $\{ \cos^2(t), \sin^2(t) \} $ is a spanning set and linearly independent then it is a basis by definition.
 A: HINT
Notice that $\cos(2t) = \cos^{2}(t) - \sin^{2}(t)$. Therefore $f\in W$ iff
\begin{align*}
f(t) & = a\cos^{2}(t) + b\sin^{2}(t) + c\cos(2t)\\\\
& = (a + c)\cos^{2}(t) + (b - c)\sin^{2}(t)\\\\
& = A\cos^{2}(t) + B\sin^{2}(t) 
\end{align*}
where $A\in\mathbb{R}$ and $B\in\mathbb{R}$.
Now it remains to prove that $\cos^{2}(t)$ and $\sin^{2}(t)$ are LI.
Can you take it from here?
EDIT
To prove that $\cos^{2}(t)$ and $\sin^{2}(t)$ are LI, we shall consider the linear combination $A\cos^{2}(t) + B\sin^{2}(t)$ and prove that it equals the null function iff $A = B = 0$.
Indeed, this is the case. To prove so, we can choose $t = 0$ and $t = \pi/2$, whence we get:
\begin{align*}
\begin{cases}
A\cos^{2}(0) + B\sin^{2}(0) = 0\\\\
A\cos^{2}(\pi/2) + B\sin^{2}(\pi/2) = 0
\end{cases} \Rightarrow A = B = 0
\end{align*}
and we are done.
Hopefully this helps!
A: Let $W$ be the subspace of all real-valued functions spanned by $\{ \cos^2(t), \sin^2(t), \cos(2t) \} $. Find a basis for W. Find the dimension of W.
Let $f$ $\in$ $W$. Since $W$ is spanned by the set $\{ \cos^2(t), \sin^2(t), \cos(2t) \}$  therefore any vector of $W$ can be written as a linear combination of these three. So for some $a, b, c \in \mathbb{R}$, one have
$f(t)=a\cos^2(t)+b\sin^2(t)+c\cos(2t)$
By the trigonometric identity, $\cos(2t)=\cos^2(t)-\sin^2(t)$,
$f(t)=a\cos^2(t)+b\sin^2(t)+c\cos(2t)$
$f(t)=a\cos^2(t)+c\cos^2(t)+b\sin^2(t)-c\sin^2(t)$
$f(t)=(a+c)\cos^2(t)+(b-c)\sin^2(t)$
$f(t)=b_1\cos^2(t)+b_2\sin^2(t)$
where $b_1$ and $b_2 \in \mathbb{R}. $
Due to the fact that $a\cos^2(t)+b\sin^2(t)+c\cos(2t)$ can be written as $b_1\cos^2(t)+b_2\sin^2(t)$ for some $b_1$ and $b_2$, $\{ \cos^2(t), \sin^2(t) \} $ is a spanning set.
Now, is it linearly independent? If $c_1\cos^2(t)+c_2\sin^2(t)=0,$ can there be  $c_1=c_2=0$?
Let $c_1$ and $c_2 \in \mathbb{R}$, $c_1=c_2$, and $c_1, c_2$ $<0$
If $c_1=c_2=2$
$\cos^2(t)+\sin^2(t)=2$
Since $c_1\cos^2(t)+c_2\sin^2(t)=0$ is only true when $c_1=c_2=0$ then it is linearly independent.
Because $\{ \cos^2(t), \sin^2(t) \} $ is a spanning set and linearly independent then $\{ \cos^2(t), \sin^2(t) \} $ is a basis for $W$ and $\dim W=2$.
