<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] As Mohamed pointed out in his answer, we can use the angle addition formula for $\sin$, viz.:
$\sin (x + \alpha) = \cos \alpha \sin x + \sin \alpha \cos x, \tag{1}$
$\sin (x + \beta) = \cos \beta \sin x + \sin \beta \cos x, \tag{2}$
$\sin (x + \gamma) = \cos \gamma \sin x + \sin \gamma \cos x, \tag{3}$
to see that all three functions lie in the subspace generated by $\sin x$ and $\cos x$, which is of course $\text{Span} \{ \sin x, \cos x \}$, and since $\sin (x + \alpha)$, $\sin (x + \beta)$, $\sin (x + \gamma)$ are thus three vectors in a subspace of dimension two, a linear dependence between them must exist.
Going a step further, we can actually solve for $\sin x$, $\cos x$ in terms of $\sin (x + \alpha)$, $\sin(x + \beta)$ and thus find the linear dependence explicitly.  We have, from 
(1) and (2),
$\begin{bmatrix} \cos \alpha & \sin \alpha \\ \cos \beta & \sin \beta \end{bmatrix} \begin{pmatrix} \sin x \\ \cos x \end{pmatrix} = \begin{pmatrix} \sin (x + \alpha) \\ \sin(x + \beta) \end{pmatrix}; \tag{4}$
the inverse of 
$\begin{bmatrix} \cos \alpha & \sin \alpha \\ \cos \beta & \sin \beta \end{bmatrix} \tag{5}$
is readily seen to be
$(\cos \alpha \sin \beta - \sin \alpha \cos \beta)^{-1}\begin{bmatrix} \sin \beta & -\sin \alpha \\ -\cos \beta & \cos \alpha \end{bmatrix} = (\sin(\beta - \alpha))^{-1}\begin{bmatrix} \sin \beta & -\sin \alpha \\ -\cos \beta & \cos \alpha \end{bmatrix}, \tag{6}$
from which it follows that
$\begin{pmatrix} \sin x \\ \cos x \end{pmatrix} = (\sin(\beta - \alpha))^{-1}\begin{bmatrix} \sin \beta & -\sin \alpha \\ -\cos \beta & \cos \alpha \end{bmatrix}\begin{pmatrix} \sin (x + \alpha) \\ \sin(x + \beta) \end{pmatrix}. \tag{7}$
(7) gives $\sin x$, $\cos x$ in terms of $\sin(x + \alpha)$, $\sin(x + \beta)$; the resulting formulas may then be inserted into (3) to express $\sin(x + \gamma)$ in terms of $\sin(x + \alpha)$, $\sin(x + \beta)$; I leave the necessary algebra to my readers.  One must of course remain vigilant to the possibility that $\sin(\beta - \alpha) = 0$; but this
is a special case best handled on its own.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
</s>