<s>[INST] Prove that $\sin(x + \alpha)$, $\sin(x + \beta)$ and $\sin(x + \gamma)$ are linearly dependent. 
Functions $f$ and $g$ are independent on an interval $D$ if $af(x) + bg(x) = 0$ implies that $a = 0$ and $b = 0$ $\forall x \in D$
let $\alpha$, $\beta$, $\gamma$ be real constants. Prove that 
$\sin(x + \alpha)$, $\sin(x + \beta)$ and $\sin(x + \gamma)$
are linearly dependent vectors in $C^0[0, 1]$. Be convincing in your reasoning (argument)

I was researching and found Wronskian. Using the Wronskian for three functions. The determinant of $f$, $g$ and $h$ is $W(f, g, h) = $
$$
        \begin{vmatrix}
        f & g & h \\
        f' & g' & h' \\
        f'' & g'' & h'' \\
        \end{vmatrix}
$$
If $W(f, g, h) \neq 0$ then $f(x)$, $g(x)$ and  $h(x)$ are linearly independent.
If $f(x)$, $g(x)$, and $h(x)$ are linearly dependent then $W(f, g, h) = 0$
My attempt 
Let
$f(x) = \sin(x + \alpha)$, $g(x) = \sin(x + \beta)$ and $h(x) = \sin(x + \gamma)$
$W(f, g, h) =$
$$
        \begin{vmatrix}
        \sin(x + \alpha) & \sin(x + \beta) & \sin(x + \gamma) \\
        \cos(x + \alpha) & \cos(x + \beta) & \cos(x + \gamma) \\
        -\sin(x + \alpha) & -\sin(x + \beta) & -\sin(x + \gamma) \\
        \end{vmatrix}
$$
$= \sin(x + \alpha)[-\sin(x + \gamma)\cos(x + \beta) + \cos(x + \gamma)\sin(x + \beta)]  - sin(x + \beta)[-\sin(x + \gamma)\cos(x + \alpha) + \cos(x + \gamma)\sin(x + \alpha)] + \sin(x + \gamma)[-\sin(x+ \beta)\cos(x + \alpha) + \cos(x + \beta)\sin(x + \alpha)]$
$= -\sin(x + \alpha)[\sin((x + \gamma) +(x + \beta))] + \sin(x + \beta)[\sin((x + \gamma) + (x + \alpha))] - \sin(x + \gamma)[\sin((x + \beta) + (x + \alpha))] = 0$
By Wronskian, $f(x)$, $g(x)$ and $h(x)$ are linearly dependent since $W(f, g, h) = 0$
Not sure if this argument is sound?
 [/INST] A short method :
Since $$\sin(x+\alpha)=(\cos \alpha) \sin x + (\sin \alpha) \cos x$$
we have $\sin(x+\alpha) \in \text{Span}\{\sin x, \cos x\}$
This implies that the rank of the family $(\sin(x+\alpha),\sin(x+\beta),\sin(x+\gamma))$ is $2$. That means that there are linearly dependants.
</s>