Verifying Linear Independence I saw the following question from M. Artin's book, Algebra. 
I simply need to show the the functions $\cos x$, $\sin x$ and $e^x$ are independent. I have no idea how to show their independence. 
Any help is appreciated but hints are appreciated more than the answer itself. If only the method of proving that the three scalars will identically equal to $0$ is mentioned, I'll be grateful. 
 A: Functions are dependent if they are linear combinations of each other, i.e. if there exists some $\alpha, \beta,\gamma$ such that the function $f=\alpha \cos + \beta\sin + \gamma \exp$ is zero.
But what does it mean for $f$ to be "zero"? What is the zero vector in the space of all functions? Well, you probably know that that's the function that maps **all ** values of $x$ to $0$.
So, you now need to find $\alpha, \beta, \gamma$ such that $\alpha\cos x + \beta\sin x + \gamma e^x$ is true for all real values $x$!
A: From the definition for linear independence you should show that if $a,b,c\in\mathbb R$ and $a\cos x+b\sin x+c e^x\equiv 0$ then $a=b=c=0$.
Because the equality should be satisfied for all $x\in \mathbb R$, then choose for example $x=0$, $x=\pi/2$, $x=\pi$. Then look at the determinant of the corresponding homogeneous linear system for the coefficients $a,b,c$
A: You need to show that if the function $f(x) = \alpha \cos\,x + \beta \sin\,x + \gamma \exp\,x$ is $0$ for all $x$ then $\alpha = \beta = \gamma = 0$. To see that $\gamma = 0$, note that $\exp\,x$ is unbounded as $x$ tends to infinity while $\sin\,x$ and $\cos\,x$ are bounded. Given $\gamma = 0$,  to see that $\alpha = \beta = 0$, note $f(0) = \beta$ while $f(\pi/2) = \alpha$.
