What is function $\sin$ without $x$? I'm solving a problem set in ODE:

This is the first time in my life that I see such functions $\sin,\cos,\tan$ without argument $x$.
I would like to ask if $\sin,\cos,\tan$ mean $\sin x,\cos x,\tan x$ or they mean something else.
Thank you so much!
 A: Usually the convention is that if $f:X\to Y$ is a function, then $f(x)\in Y$ is the value at the point $x\in Y$. As you can do arithmetic with function values, you can do arithmetic with functions, applying the operation pointwise. Thus $h=f+g$ is $h(x)=f(x)+g(x)$ for all $x$.
This gives a little complication if you want to write equations like 1) without too much decorations like writing $\forall x\in X$. One possible interpretation would then be that $x$ has a double meaning, it is the variable name and, where appropriate, a function itself, the identity function. So that 1) would be in this convention $xy'+y=2x$, avoiding the function evaluation notation.
In 3-5, the pendulum is moving to the extreme other side regarding the function evaluation notation in that $\sin$ etc. is used as function name in the same way as $f$ above. This is a very unusual notation, but not formally wrong. But it will look very strange in larger terms, $\sin+\cos$ is tolerable, but how to interpret $\sin\cos$?
A: In this context you are dealing with functions $\mathbb R\to\mathbb R$.


*

*$x$ stands for the identity function.

*$y$ stands for some function.

*$y'$ stands for the derivative of function $y$.

*$\sin$ stands for the function prescribed by $z\mapsto\sin(z)$ (same story for $\cos$ and $\tan$).


But I must say that there is some inconsistency which causes confusion. 


*

*The author sometimes uses $y$ and another time $y(x)$. 

*If $x$ is used as a function then IMV it is better to use another symbol (e.g. $z$) as argument of the function. Another way is not writing $x$ but writing $\mathsf{id}$ so that $x$ can be reserved as argument (as we are used to).

A: A univariate function $f$ is usually defined by a domain, a codomain, and an analytic specification of how an input generates an output. The latter is given by specifying what $f(x)$ generates for any $x$ in the domain of $f$. Thus, $f$ is a function, but $f(x)$ is not: $f(x)$ is an output. Similarly, we may let $\sin$ denote the specific function that for any input $x$ in its domain, namely $\mathbb{R}$, generates the sinus of $x$, denoted as $\sin(x)$. Interpreted this way, the third equation you present can be read as: solve $y
'(x)\cos(x)-y(x)\sin(x)=-\cos(x)$ over $x$ in $\mathbb{R}$. 
A: $\sin$ is an abbreviation for $\sin(x)$, in the same manner that $y'$ is an abbreviation for $y'(x)$ and $y$ for $y(x)$.
