linear independence question I'm asked to show linear independence for the following familiy of functions as functions on the interval $[-\pi, \pi]$:
1) $A= \{f_1,\ f_2, \ f_3 \}$ where $\ f_1(x)=1$, $\ f_2(x)=\cos x$, and $\ f_3(x)=\sin x$.
It's been a long time since I prove linear independence and I'm recently trying to sharpen my linear algebra skills. Can I use wronskian?
 A: Suppose that $$\alpha+\beta\sin x+\gamma\cos x=0$$
When $x=0$, $\alpha+\gamma=0\implies \alpha=-\gamma$ so $$\alpha+\beta\sin x-\alpha\cos x=0$$
When $x=\pi$ we get $\alpha+\alpha=0$ so $\alpha=0$. And...?

Alternatively, suppose that $$\alpha+\beta\sin x+\gamma\cos x=0$$
Integrate over $[-\pi,\pi]$. You get $2\pi\alpha=0$, since the other two integrals vanish. Then multiply by $\sin x$ and integrate. Then multiply by $\cos x$ and integrate. In general, you can show any set of functions $$\{\cos mx,\sin nx:m=0,1,2,\ldots\;, n=1,2,\ldots\}$$ is linearly independent by using the orthongonality relations:
$$\int_{-\pi}^\pi \cos mx\cos nxdx=\delta_{mn}\pi$$
$$\int_{-\pi}^\pi \sin mx\sin nxdx=\delta_{mn}\pi$$
$$\int_{-\pi}^\pi \cos mx\sin nxdx=0$$
A: You can indeed, the Wronskian being the determinant
$$
\begin{vmatrix}
f_1 & f_2 & f_3 \\ 
f_1' & f_2' & f_3' \\ 
f_1'' & f_2'' & f_3''
\end{vmatrix}
=
\begin{vmatrix}
1 & \cos x & \sin x \\ 
0 & -\sin x & \cos x \\ 
0 & -\cos x & -\sin x
\end{vmatrix}
$$
If this determinant is non-zero anywhere on $[-\pi,\pi]$ then the functions are linearly independent on that interval. 
A: Suppose that $a+b\cos x +c\sin x$ is identically $0$. Differentiate  twice. We get $-b\cos x-c\sin x$ is identically $0$. It follows that $a=0$. Put $x=0$ in $-b\cos x-c\sin x=0$. We get $b=0$. It follows that  $c=0$.
Remark: The fact that differentiation is linear is often useful. 
