# Are the following functions linearly independent?

I have 4 functions.

$$f_{1} = \cos(x)$$ $$f_{2} = \sin(x)$$ $$f_{3} = e^x\cos(x)$$ $$f_{4} = e^x\sin(x)$$

Is $$\{f_{1}, f_{2}, f_{3}, f_{4}\}$$ linearly independent in $$C(\mathbb{R})$$? My inital conclusion is no because the linear combination of these functions $$af_{1} + bf_{2} + cf_{3} + df_{4} = 0$$ for $$a=b=c=d=1\neq 0$$ namely for when $$x = 3\pi/4$$. However, I'm not sure if my reasoning is correct. Any help in explaining whether it is linearly independent or dependent would be helpful. Thank you.

Hint: Linear dependence means we have $$af_1(x)+bf_2(x)+cf_3(x)+df_4 (x)=0$$ for every $$x$$ (with not all coefficients $$0$$), not just for one particular value of $$x$$.
Put $$x=0, x=\pi, x=\pi /2$$ and $$x =\pi /3$$. (Do you know the values of $$\sin x$$ and $$\cos x$$ for these values of $$x$$?). You will get $$4$$ equations for $$a,b,c,d$$. Try to show from these equations that $$a=b=c=d=0$$.
To prove that these functions aren't independent you would need to prove that there exists a nontrivial linear combination that forms the zero vector in that space. In your case, this means $$af_1 + bf_2 + cf_3 + df_4 = f_0$$ where $$f_0$$ is the constant function defined as $$f_0(x) = 0$$ for all $$x\in \mathbb{R}$$.
It is not enough to see that the resulting function is zero at a specific point such as $$x=3\pi/4$$. It must be null everywhere, since this is the property that defines the zero function in $$\mathit{C}(\mathbb{R})$$.