Gfrktyl

Let $mathbb{R}^mathbb{R} $ be a vector-space of functions $f:mathbb{R}→mathbb{R}$.
The following functions in the vector space are defined as follows:
$$f(x) = sin(x),qquad g(x) = cos(x),qquad h(x) = x.$$
Is the triple $(f,g,h)$ linearly dependent?

I have a feeling it is, because $h(x)$ is the identity function, so $h(sin(x)) = sin(x)$, and I know the definitions, I can't figure out how to put my explanation (if it is correct) properly.

Thanks in advance.

Your feeling is not correct. Recall that, by definition, if such functions are linearly dependent then there are real numbers $A,B,C$ not all zero such that
$$Asin(x)+Bcos(x)+Cx=0 quad forall xin mathbb{R}.$$
Now by letting $x=0,pi/2,pi$, we have three linear equations
$$begin{cases}
Acdot 0+Bcdot 1+Ccdot 0=0qquad &(x=0)\
Acdot 1+Bcdot 0+Ccdot frac{pi}{2}=0qquad &(x=frac{pi}{2})\
Acdot 0+Bcdot (-1)+Ccdot pi=0qquad &(x=pi)\
end{cases}$$
What may we conclude about $A,B,C$? Can you take it from here?

Asserting that ${f,g,h}$ is linearly independent means that$$(forall a,b,cinmathbb{R}):af+bg+ch=0implies a=b=c=0.$$If $af+bg+ch=0$, then $af(0)+bg(0)+ch(0)=0$, which means that $b=0$. You still have $af+ch=0$. But then $afleft(fracpi2right)+cfracpi2=0$, which means that $a+cfracpi2=0$, and $af(pi)+ch(pi)=0$, which means that $cpi=0$. So, $a=c=0$ too.