Are $\sin x$, $\sin(x + \frac{\pi}{6})$, and $\sin(x + \frac{\pi}{3})$ linearly independent? I'm specifically interested in the dimensionality of the subspace spanned by $\{\sin x, \sin(x + \frac{\pi}{6}), \sin(x + \frac{\pi}{3})\}$ in $C^0(\mathbb{R}, \mathbb{R})$, the vector space of continuous functions from $\mathbb{R}$ to $\mathbb{R}$.
I think the answer is, yes, they are linearly independent, but I'm not confident about the best way to demonstrate this. Would a good approach be to do Taylor expansions at $x_0 \in \{0, \frac{\pi}{6}, \frac{\pi}{3}\}$, and examine the polynomial coefficients? This seems messy. Wondering if there's a simpler way to think about this.
 A: Using the expansion of $\sin(a+b)$ we get:
\begin{eqnarray*}\sin\left(x+\frac\pi 6\right)&=&\sin x \cos\frac\pi6+\cos x \sin\frac\pi6\\&=&\frac{\sqrt{3}}2\sin(x)+\frac12\cos x\end{eqnarray*}
\begin{eqnarray*}\sin\left(x+\frac\pi 3\right)&=&\sin x \cos\frac\pi3+\cos x \sin\frac\pi3\\&=&\frac{1}2\sin(x)+\frac{\sqrt{3}}2\cos x\end{eqnarray*}
Thus $$\sqrt{3}\sin\left(x+\frac\pi 6\right)-\sin\left(x+\frac\pi 3\right)=\sin(x)$$
A: First, note that $\{ \sin x, \cos x \}$ is linearly independent.  Let $f(x)= A \sin x + B \cos x$ with $B \neq 0$.  Then $f(x)= \sqrt {A^2+B^2} \sin (x+ \theta)$ for $\theta= \dfrac{B}{\vert B \vert} \cos^{-1} \left ( \dfrac{A}{\sqrt {A^2+B^2}}\right )$, which is not the $0$ function, and if $B=0, f(x)=A \sin x,$ which is only the zero function if also $A=0$.
Now use angle-addition formulas to convert $\sin (x+ \frac {\pi}{6})$ and $\sin (x + \frac{\pi}{3})$ to a linear combination of $\sin x$ and $\cos x$.  Since $\sin x$ and $\cos x$ span a subspace of dimension $2$ and we have just shown that your set of three functions lies within that subspace, your set cannot be linearly independent.
A: Your functions are linearly dependent. To see this, it suffices to prove that $e^{ix} , e^{i(x+ \pi/6)}, e^{i(x+ \pi/3)}$ are linearly dependent over $\mathbb{R}$. Taking imaginary parts will solve your problem.
We need to find real numbers $A,B,C \neq 0$ such that
\begin{align*}
0 & = Ae^{ix} + Be^{i(x+ \pi/6)} + Ce^{i(x+ \pi/3)} \\
 & = e^{ix} ( A + e^{i\pi/6}B + e^{i\pi/3}C ) 
\end{align*}
Since $e^{ix}\neq 0$ for any $x$, we must find $A,B,C$ such that $A + e^{i\pi/6}B + e^{i\pi/3}C = 0$. 
Here's one way we can do this:
Pick any $C$. Then, choose $B$ so that  $\text{Im}(e^{i\pi/6}B + e^{i\pi/3}C) =0$. In other words, choose $B$ so that $e^{i\pi/6}B + e^{i\pi/3}C$ is real. Then choose $A$ to be the negative of that so that the entire sum is $0$.
