Prove that the set $ \{\sin(x),\cos(x),\sin(2x),\cos(2x)\}$ is linearly independent. Prove that the set $ \{\sin(x),\cos(x),\sin(2x),\cos(2x)\}$ is linearly independent.

I have tried plugging in values for $x$ but where does this lead? I know what the Wronksian is and it would give an almost trivial solution in this case, but I am not allowed to use that.
 A: Let $a,b,c,d\in\mathbb{R}$ such that $x\mapsto a\sin x + b\cos x+ c\sin 2x + d\cos 2x$ is the $0$ function. You have to show that one must have $a=b=c=d=0$.
For $x=0$, we get $b+d=0$. For $x=\pi$, $-b+d=0$. Combining the two, $b=d=0$.
Now, you have $a\sin x + c\sin 2x = 0$ for all $x$. Can you use similar techniques (or new ones -- differentiating and then using the same technique, for instance) to show that $a=c=0$ as well? (even without using differentation, just considering $x=\frac{\pi}{2}$ may help).
A: suppose $$a \cos x + b \sin x + c \cos 2x + d \sin 2x = 0$$ for all $x.$ putting $x = 0$ gives $$a + c = 0 \tag 1 $$
put $x = \pi$ gives $$ -a +c= 0 \tag 2$$
$(1)$ and $(2)$ shows $a = 0, c =0.$
put $x = \pi/2$ gives $ b = 0 $ now we are left with $d \sin 2x = 0$ which implies $d = 0.$
so we have shown that $a \cos x + b \sin x + c \cos 2x + d \sin 2x = 0$ implies $a=b=c=d=0$ which means $\{\cos x,\sin x,  \cos 2x,\sin 2x  \}$ is linear independent.
A: My answer is based on fact that orthogonality implies linear independence.
Let $S = \{\sin(x),\cos(x),\sin(2x),\cos(2x)\}$
and $f(x),g(x) \in S$
Then $\int_{0}^{2\pi} f(x)g(x)dx = 0$ for $f(x) \neq g(x)$
Hence $S$ forms an orthogonal set and thus $S$ is a linearly independent set.
A: To show that
$\sin(kx), \cos(kx)$
for $k = 1, ..., n$
are linearly independent,
assume they are not,
so
$0
=\sum_{k=1}^n (a_k \sin(kx)+b_k \cos(kx))
$.
Now use the orthogonality relations for
$\sin$ and $\cos$:
to show that
 $a_k = 0$,
multiply by
$\sin(kx)$
and integrate from
$-\pi$ to $\pi$;
all the terms except that
integrate to zero, 
and the term with
$a_k$ does not,
so we get
$a_k = 0$.
Similarly,
to show $b_k = 0$,
multiply by
$\cos(kx)$
and integrate.
This property is useful.
I did not discover it.
