I have problems finding a dual basis. (exercise b)
$B:=(\sin(x),\cos(x)) \in \mathbb{R}^\mathbb{R}$ is a basis of a subspace spanned by sine and cosine. $V:=\langle\sin(x),\cos(x)\rangle \subseteq \mathbb{R}^\mathbb{R}$.
Let $(\alpha, \beta) \in (V^*)^2$ with
$\alpha: V\rightarrow \mathbb{R}: p \mapsto p(\pi/2)$ and
$\beta: V\rightarrow \mathbb{R}: p \mapsto p(0)$
Show that:
a) $\alpha$ and $\beta$ are linear functionals
b) $(\alpha, \beta)$ is dual basis of $B$.
My approach:
Since $p \in V$, p must have the form $ p = a\cdot\sin(x)+b\cdot\cos(x)$.
So for $\alpha$, p maps to $a\cdot\sin(\pi/2)+b\cdot\cos(\pi/2) = a$ and for $\beta$, p maps to $a\cdot\sin(0)+b\cdot\cos(0) = b$.
a)
$\alpha$ is a linear functional since $p(\pi/2)(k\cdot(x+y)) = a\cdot(k\cdot(x+y)) =ka(x)+ka(y)= ka\cdot p(\pi/2)(x)+ka\cdot p(\pi/2)(y)$
$\beta$ is a linear functional since $p(0)(k\cdot(x+y)) = b\cdot(k\cdot(x+y)) =kb(x)+kb(y)= kb\cdot p(0)(x)+kb\cdot p(0)(y)$
b)
For a dual basis I need that $\alpha(\sin(x)) = 1, \alpha(\cos(x)) = 0, \beta(\sin(x)) = 0, \beta(\cos(x)) = 1$.
But with my $\alpha$ and $\beta$ it would not work. So where is my mistake?