Linear independence of the functions $1,\cos(x),\cos(2x)$ I want to show that the functions $1,\cos(x),\cos(2x)$ are linearly independent in $C[-\pi,\pi]$. I computed the Wronskian determinat of these functions but at the points $x=0,-\pi,\pi$ the obtained determinant vanishes. So, I couldn't obtained the required result. Can you me help me in showing that these are linearly dependent or linearly independent in $C[-\pi,\pi]$.
Thanks in advance. 
 A: Suppose $a+b\cos x + c\cos 2x = 0$ for all $x \in [-\pi,\pi]$, where $a,b,c$ are real constants. 
$x = 0$ yields $a+b+c = 0$, 
$x = \pi/2$ yields $a-c = 0$, 
$x = \pi$ yields $a-b+c = 0$
Are there any solutions besides $a = b = c = 0$?
A: Hint: Linear dependency in $C[-\pi,\pi]$ would require a non-trivial triple $(c_1,c_2,c_3)$ of constants such that the equation
$$
c_1\cdot1+c_2\cdot\cos x+c_3\cdot\cos2x=0\qquad(*)
$$
holds for all $x\in[-\pi,\pi]$. If that is the case, then the equation $(*)$ holds for all $x\in(0,\pi)$, too. Your Wronskian calculation on the other hand ...
A: you should prove that this functions are orthogonal so you can use dot multiply of functions and prove that is 0. 

A: Suppose that $a+b\cos x+c\cos 2x=0$ in our interval. Differentiate and set $x=\pi/2$. We get $b=0$.
So $a+c\cos 2x=0$. Set $x=\pi/4$. We get $a=0$.
A: Try a different value of $x$ in your Wronskian.  Hint.  The second row (or column, depending how you set it out) of the Wronskian is
$$0\ ,\quad -\sin x\ ,\quad -2\sin2x\ :$$
taking $x$ to be a multiple of $\pi$ will make this $(0,0,0)$ which is of no use to you.  So try something other than a multiple of $\pi$.
