linear independence with $\sin x, \cos x$ I don't know why $\sin x$ and $\cos x$ are lineary independent since if we take linear combination $a\cdot \sin x + b \cdot \cos x=0$ and for $a=\sqrt{3}$ and $b=1$ and $\displaystyle x=\frac{\pi}{6}$ we have that
$a\cdot \sin x + b \cdot \cos x=0$ and $a \neq 0 $ and $b \neq 0$ so it suggest they aren't independent
 A: The problem is when you take $x=\pi/6$.  It is true that the values of the function at $x=\pi/6$ are not linearly independent, and that's what you've shown here.  They are in the space of all real numbers, and that is a one-dimensional space, so more than one element cannot be linearly independent.
For functions of $x$ to be linearly independent, the equality has to hold for all values of $x$, not just for some.
A: They are linearly independent in a type of vector space called a function space. In this vector space the zero vector is the function $f(x) = 0$. To say that $\sin x$ and $\cos x$ are linearly independent is to say that no linear combination of $\sin x$ and $\cos x$ results in the zero function.
A: By the same logic, you would argue that $1$ and $x$ are linearly dependent since $a \cdot 1 + b \cdot x = 0$ if $a = b = 1$ and $x = -1$.
The catch here is that two functions are linearly dependent iff there is a nontrivial way to form a linear combination in order to obtain the zero function (which is zero for any possibly choice of $x$).
A: For two functions $f$, $g$ to be linearly independent, you need a nontrivial linear combination to be the zero-function.  That is, you'd need $a,b\in\mathbb{R}$, not both $0$, so that $a\sin(x)+b\cos(x)=0$ for all $x$.
This can't happen, because checking the equation at $x=0$ and $x=\frac{\pi}{2}$ shows $a$ and $b$  must both be $0$.
A: $f(x),g(x)$ are linear independent if there exists $a,b \in \mathbb{R}$ that $a\neq 0$ or $b\neq 0$ for $\textbf{all}$ there is:
$$af(x)+bg(x)=0$$
Suppose that $f(x)=\sin(x)$ and $\cos(x)$ are linear dependent. Take $x_1=0$ and $x_2=\frac{\pi}{2}$, you get $a=b=0$.
