Linear Algebra: coordinates and basis I'm asked to show that if $f,g \in \mathrm{Sig}_3$ (sig stands for signal) have the coordinates $x,y \in \mathbb{R^4}$ with respect to the basis $(v_0, v_1, v_2, v_3)$, then 
$$\left<f,g\right>=\pi(x_1y_1+x_2y_2+x_3y_3+x_4y_4)$$
where
$$v_0=\frac{1}{\sqrt2}, 
v_1=\cos x,
v_2=\cos(2x),
v_3=\cos(3x)$$
and
$$C^0([0;2\pi])$$
Any idea how to show this? I'm completely stuck. All of the examples online use vectors with numbers, but I'm just given functions.
 A: Assume $f=[a_1,a_2,a_3,a_4]$ and $g=[b_1,b_2,b_3,b_4]$, you'll have $$f(x)=a_1\frac1{\sqrt2}+a_2\cos x+a_3\cos (2x)+a_4\cos (3x)\\g(x)=b_1\frac1{\sqrt2}+b_2\cos x+b_3\cos (2x)+b_4\cos (3x)$$
The inner product $\left<f,g\right>$ would be
$$\left<f,g\right>=\int_0^{2\pi}f(x)g(x)\mathrm{d}x$$ Before we begin integrating it's worth noting that for every distinct $m,n \in\Bbb{Z}$ and $m,n\geq0$
$$
\begin{align}
\int_0^{2\pi}\cos(mx)\cos(nx)\mathrm{d}x&=\frac12\int_0^{2\pi}\cos((m+n)x)+\frac12\int_0^{2\pi}\cos((m-n)x)\mathrm{d}x\\
&=0+0\\&=0
\end{align}
$$
(Note that the RHS integrals are on a whole period of $\cos({kx})$ where $k\in\Bbb{N}$, so they equal zero.)
Going back to the inner product, now we know that the integrals $\int_0^{2\pi}\cos x\mathrm{d}x$, $\int_0^{2\pi}\cos(2x)\mathrm{d}x$, $\int_0^{2\pi}\cos(3x)\mathrm{d}x$, $\int_0^{2\pi}\cos x\cos(2x)\mathrm{d}x$, $\int_0^{2\pi}\cos x\cos(3x)\mathrm{d}x$, and $\int_0^{2\pi}\cos(2x)\cos(3x)\mathrm{d}x$ are zero, so we can omit them from the multiplication result. What remains is
$$
\begin{align}
\left<f,g\right>&=\int_0^{2\pi}f(x)g(x)\mathrm{d}x\\
&=\int_0^{2\pi}\left(a_1b_1\frac12+a_2b_2\cos^2x+a_3b_3\cos^2(2x)+a_4b_4\cos^2(3x)\right)\mathrm{d}x
\end{align}
$$
but since for $k\in\Bbb{Z}$
$$
\begin{align}
\int_0^{2\pi}\cos^2(kx)\mathrm{d}x&=\int_0^{2\pi}\left(\frac12+\frac12 \cos(2kx)\right)\mathrm{d}x\\
&=\pi+0\\&=\pi
\end{align}
$$
we'll have
$$
\begin{align}
\left<f,g\right>&=a_1b_1\pi+a_2b_2\pi+a_3b_3\pi+a_4b_4\pi\\
&=\pi(a_1b_1+a_2b_2+a_3b_3+a_4b_4)
\end{align}
$$
which is what you want.
A: The question is a bit hard to understand because it's written in a weird order.  Usually the question goes (environment) (specifics) (statement to be proved).  If I understand your problem well, I think it would be better written this way:

Let $f$ and $g$ be elements of $\mathrm{Sig}_3$, which has inner product
  $$
    \left<f,g\right> = \int_0^{2\pi} f(x)g(x)\,dx
$$
  [I don't actually know what $\mathrm{Sig}_3$ is, even with the information that Sig stands for signal.  But I'm trying to work from the context.]  Let
  \begin{align*}
   v_0 &= \frac{1}{\sqrt{2}} \\
   v_1 &= \cos(x) \\
   v_2 &= \cos(2x) \\
   v_3 &= \cos(3x)
\end{align*}
  Suppose that with respect to the basis $(v_0, v_1, v_2, v_3)$, $f$ and $g$ have coordinates $x$ and $y$ in $\mathbb{R}^4$.  Show that
  $$
    \left<f,g\right> = \pi(x_1 y_1 + x_2 y_2 + x_3 y_3 + x_4 y_4)
$$

Please correct me if that's not what your problem is.  But notice how it starts with the environment of the vector space $\mathrm{Sig}_3$, along the definition of the inner product.  Then comes the specific information: for these two elements, we have these two coordinate vectors.  Finally, the statement we are to prove.
So I think you will be home free if you understand the meaning of this sentence:

Suppose that with respect to the basis $(v_0, v_1, v_2, v_3)$, $f$ and $g$ have coordinates $x$ and $y$ in $\mathbb{R}^4$.

What does this tell you about the formulas for $f(x)$ and $g(x)$?  Once you know this, you should be able to compute $\left<f,g\right>$ directly.
