Linear independence of $\sin(x)$, $\sin(2x)$, $\sin(3x)$ in Map($\mathbb{R},\mathbb{R}$) It seems rather obvious to me that $\sin(x)$, $\sin(2x)$, $\sin(3x)$ are linearly independent in $\operatorname{Map}(\mathbb{R},\mathbb{R})$, but I'm not sure how to prove it (or disprove it if I'm wrong?).
I know that for linear independence it must
$$ a\sin(x)+b\sin(2x)+c\sin(3x)=0 $$
$$ \implies a=b=c=0$$
but how can I show this is true for any $x \in \mathbb{R}$?
 A: Hint: if $f(x) = a\sin (x) + b\sin(2x) + c\sin(3x) = 0$
then compute $$
\int _0^{2\pi} f(x) \sin(x)dx\\
\int _0^{2\pi} f(x) \sin(2x)dx\\
\int _0^{2\pi} f(x) \sin(3x)dx\\
$$
A: Differentiate the LHS of the equality and evaluate at $0$ we get
$$a+2b+3c=0\tag1$$
Now differentiate twice we get
$$a+2^3b+3^3c=0\tag2$$
and again we get
$$a+2^5b+3^5c=0\tag3$$
and finally $(1),(2),(3)$ give the desired result.
A: Pick three values for $x$, for example $\pi/2,\pi/3,\pi/4$. The first gives $a = c$. The last gives you $b = 0$. The second will give you $a=c=0$.
A: Let $a,b,c\in\mathbb R$ such that $a\sin(x)+b\sin(2x)+c\sin(3x)=0$ for all $x\in\mathbb R$. Denote $f(x)=a\sin x+b\sin 2x+c\sin 3x$. Then $f'(x)=0$ and $f''(x)=0$ for all $x\in\mathbb R$. Hence, for any $x\in\mathbb R$ for $a,b,c$ we have a homogeneous system of linear equations ($a,b,c$ - unknowns)
\begin{cases}
a\sin(x)+b\sin(2x)+c\sin(3x)=0\\
a\cos(x)+b 2\cos(2x)+c 3\cos(3x)=0\\
a\sin(x)+b 4\sin(2x)+c 9\sin(3x)=0
\end{cases}
with a matrix
$$
W(x)=
\begin{pmatrix}
\sin(x) & \sin(2x)  & \sin(3x)\\
\cos(x) & 2\cos(2x) & 3\cos(3x)\\
\sin(x) & 4\sin(2x) & 9\sin(3x)
\end{pmatrix}.
$$
Obviously,
$$
\det(W(\frac{\pi}{2}))=
\begin{vmatrix}
1  &  0  & -1\\
0  & -2  &  0\\
1  &  0  & -9
\end{vmatrix}\neq 0,
$$
therefore our system has only zero solution, i.e. $a=b=c=0$.
A: The hypothesis is that $a\sin(x)+b\sin(2x)+c\sin(3x)=0$ for all $x\in\mathbb{R}$, because this should be the zero function.
Now you can evaluate the expression for any particular value of $x$; if we do it for $x=\pi/2$ we get
$$
a\sin\frac{\pi}{2}+b\sin\pi+c\sin\frac{3\pi}{2}=0
$$
that is, $a-c=0$.
Now we can try $x=\pi/4$, that gives
$$
a\sin\frac{\pi}{4}+b\sin\frac{\pi}{2}+c\sin\frac{3\pi}{4}=0
$$
or $a/\sqrt{2}+b+c/\sqrt{2}=0$.
Plug in another suitable value and you'll be done.
