Is $\{\sin x,\cos x\}$ independent? Is $\{\sin x,\cos x\}$ linearly independent in $\mathbb{R}^n$?
I thought they were not because I can write $\cos x=\sin (x+\pi/2)$.
My professor on the other hand said it was independent and his proof is as follows:

If $\{\sin x,\cos\}$ is independent in $[0,2\pi]$, then it will be independent on all of $\mathbb{R}.$ (He didn't prove this either.)
$a\cos x+b \sin x=0  ,\forall \vec{x}\in[0,2\pi].$
$x=0\implies a\cdot1+b\cdot0=0 \implies a=0$.
$x=\pi /2\implies a\cdot 0+b \cdot 1 \implies b=0$.
So $\{\sin x, \cos x\}$ is independent on $[0,2\pi],$ and thus independent everywhere.
$QED$.

Is there something I am missing or not understanding...? Why is this set independent, when I can express an element of the set as a linear combination?
 A: You proved that there were no non-trivial (ie, with at least one non-zero coefficient) linear combination of $\sin$ and $\cos$ identically zero on $[0,2\pi]$. That by definition means that $\{\sin,\cos\}$ is linearly independent on $[0,2\pi]$.
This also implies that they are on $\mathbb{R}$, because in particular any identically zero linear combination on $\mathbb{R}$ is also zero on $[0,2\pi]$ (the reciprocal is also true, actually, by $2\pi$-periodicity).
As for your initial comment about $\cos x = \sin(x+\frac\pi 2)$ for all $x$, it is not a linear combination. You are not writing $\cos=\alpha\sin$ for some scalar $\alpha$, you are writing $\cos=\sin\circ f$ for some function $f$.
A: I am surprised no one has yet pointed out that your question

Is $\{\sin x,\cos x\}$ linearly independent in $\mathbb{R}^n$?

doesn't make sense.
For a start, $\sin x$ and $\cos x$ aren't in $\mathbb{R}^n$! Your notation is a little ambiguous: you could mean the functions $\sin$ and $\cos$ (in which case they live inside a vector space of functions), or you could mean the numbers $\sin x$ and $\cos x$ for some fixed value $x$, say $x = 5$ (in which case they live inside a vector space of numbers, like $\mathbb{R}$).
I'm going to assume you are thinking of them as functions. So let's be clear about this: let's define $V$ to be the $\mathbb{R}$-vector space spanned by the two vectors $\mathbf{u}$ and $\mathbf{v}$, where secretly $\mathbf{u}$ is the function $\mathbf{u}(x) = \sin x$ and $\mathbf{v}$ is the function $\mathbf{v}(x) = \cos x$.
Now, $V$ is a vector space of functions - what's the zero vector? Of course, it's the zero function, $\mathbf{0}$, defined by $\mathbf{0}(x) = 0$. What does linear independence mean? Well, it means that we can find real numbers $a$ and $b$ such that $a\mathbf{u} + b\mathbf{v} = \mathbf{0}$. But these are functions, and two functions $f$ and $g$ are equal when they agree at all $x$, i.e. $f(x) = g(x)$ for all $x$.
That is, we want $a\mathbf{u}(x) + b\mathbf{v}(x) = 0$ for all $x$, or in more familiar language, $a\sin x + b\cos x = 0$ for all $x$. Now, what are $a$ and $b$?
A: 
By definition $\{\cos(x), \sin(x)\}$ is a set of linearly independent in $C(\mathbb{R})$ if for all $a,b\in\mathbb{R}$ that satisfies the equation 
  $$
a\cos(x)+ b\sin(x)=0 
$$
  for all $x\in\mathbb{R}$  implies $a=0$ and $b=0$. 

Supose that $a\neq 0$ or $b\neq 0$ and
$
a\cos(x)+ b\sin(x)=0
$
for all $x\in\mathbb{R}$. Check that for values ​​of $ a $ and $ b $ arbitrary but fixed the equality hold only for fixed values ​​of $ x $ and not all values ​​of $ x $ in $ \mathbb{R} $. And this contradicts our initial assumption.
A: Your professor's first claim (that we just have to show linear independence of $\sin x$ and $\cos x$ on $\lbrack0,2\pi\rbrack$) follows from $\sin x$ and $\cos x$ both being $2\pi$ periodic; two real numbers $a$ and $b$ satisfy $a \sin\theta + b\cos\theta = 0$ for all $\theta \in \mathbb{R}$ if and only if $a \sin x + b \cos x = 0$ for all $x \in \lbrack 0, 2\pi\rbrack$. To prove this, note $\implies$ is clear. For $\Longleftarrow$, note every $\theta \in \mathbb{R}$ is of the form $\theta = x + 2k\pi$ for some $x \in \lbrack0,2\pi\rbrack$ and $k \in \mathbb{Z}$, and so $a \sin x + b \cos x = 0$ and $2\pi$ periodicity imply $a\sin\theta + b\cos\theta = a \sin(x + 2k\pi) + b\cos(x + 2k\pi) = 0$.
For your other question, the statement $\cos x = \sin(x + \pi/2)$, $\forall x \in \mathbb{R}$ is an equation of linear dependence, but not for the vectors/functions $f(x) = \cos x$ and $g(x) = \sin x$. This is because $g(x)$ isn't in your equation: $h(x) = g(x+\pi/2) \sin(x+\pi/2)$ is! In fact, what you've shown is that $\{f(x),h(x)\} = \{\cos x, \sin(x + \pi/2)\}$ is a linearly dependent set.
I can elaborate if you still have questions.
