vector space without verification of the axioms I'm trying to show that the functions $c_1 + c_2 \sin^2 x + c_3 \cos^2 x$ forms a vector space.
And I will need to find a basis of it, and its dimension.
Is there a way how to do this without verifying the 8 axioms for a vector space, and if we let the set $X = \{c_1 + c_2 \sin^2 x + c_3 \cos^2 x\}$ then we note that $1 = \sin^2 x + \cos^2 x$, and this is enough. So the dimension is $2$.
Thanks.
Can you please provide clarification on how the argument of the subspace of the vector space follows? I think you did it already by inspection, but its not very complete to me, can you please write it down? Thanks
 A: Yes, you can show that it's a subspace of some other vector space. Letting $$V=\{c_1 + c_2 \sin^2(x) + c_3 \cos^2(x)\,\vert\,c_i\in \mathbb{R}\},$$ and letting $W$ be the space of all functions from $\mathbb{R}$ to $\mathbb{R}$ (under the operations of point-wise addition and scalar multiplication), it is clear that $V\subseteq W$. 
Now, all that you need to do is show that for all $\alpha,\beta\in V$ and all $a,b\in \mathbb{R}$, that we have $a\alpha + b \beta\in V$. That's simple enough that it practically writes itself.
(Note: I'm assuming that the underlying field is $\mathbb{R}$, as it usually is for an undergraduate-level linear algebra course.)
A: Every set $X$ of objects forms a vector space over chosen field.
It is the free vector space over $X$: $F(X)$. Its basis is the set $X$ and its dimension is the cardinality of $X$.
You must be careful what do you mean by saying that something forms something.
A: To prove that the span of a set of vectors forms a subspace of a vector space one can use the subspace test theorem. Suppose $W = span\{ v_1,v_2, \dots v_k \}$ where "span" means the set of all linear combinations with coefficients from $\mathbb{R}$. Note $W \neq \emptyset$ as $0$ is a linear combination $0v_1+0v_2+ \cdots +0v_k=0$. Moreover, if $x,y \in W$ and $c \in \mathbb{R}$ then $x = x_1v_1+ \cdots x_kv_k$ and $y=y_1v_1+\cdots +y_kv_k$ for some real constants $x_i,y_j \in \mathbb{R}$. Consider then:
$$ cx+y = c[x_1v_1+ \cdots x_kv_k]+y_1v_1+\cdots +y_kv_k = (cx_1+y_1)v_1+\cdots + (cx_k+y_k)v_k $$
Thus $cx+y \in W$. It follows that the nonempty $W$ is closed under scalar multiplication and vector addition and by the subspace test we find $W$ is a subspace. This means $W$ is a vector space with respect to the operations of the vector space $V$ which contains $W$.
Now, you can take the redundant set $\{ 1, \cos^2 \theta, \sin^2 \theta \}$ as a spanning set for your subspace $W$, however this would not be a basis.
To find a basis you need to select linearly independent vectors whose span is $W$.
You already pointed out $\sin^2 \theta+\cos^2 \theta=1$ in your post. Think about this. You can see how to write one of the vectors in $\{ 1, \cos^2 \theta, \sin^2 \theta \}$ as a linear combination of the remaining vectors. You have at least three obvious choices for the basis here. Hope this helps.
