This is a mathematical logic question.
"All men are mortal" seems to be represented as ∀x (Man(x) ⟶ Mortal(x)). "Some trigonometric functions are periodic" is represented as ヨx(Trigonometric(x) ⋀ Periodic(x)). Why isn't the latter represented as ヨx(Trigonometric(x) ⟶ Periodic(x))? Or the former using conjunction??
The expression $A \implies B$ does not mean "$A$ is $B.$" It does not even mean "if $A$ then $B$" in ordinary English, because ordinary English is ambiguous about what that sentence means, whereas $A \implies B$ is not ambiguous. A more accurate translation into ordinary English is, "Either $B$ is true, or $A$ is false, or both."
Now if we believe the assertion that no matter who or what we choose to talk about about, either he/she/it is mortal or he/she/it is not a man, then we believe the conventional meaning of the statement "all men are mortal."
On the other hand, if we write $\exists f(\mathrm{Trigonometric}(f)\implies\mathrm{Periodic}(f)),$ then we have just asserted that there is at least one function $f$ such that either $f$ is periodic or $f$ is not a trigonometric function. And that statement is very easily shown true simply by exhibiting a function that is not trigonometric; for example, take $f(t) = t,$ which is neither trigonometric nor periodic, but since it satisfies the "not trigonometric" clause of the disjunction, it makes the entire disjunction true.
But that is not at all the same thing as saying some trigonometric functions are periodic. Our statement doesn't actually tell us very much about either trigonometric functions or periodic functions. All it does is rule out the possibility that every single function that can ever exist is a non-periodic trigonometric function, and that no other kinds of functions exist.
In general, in universal quantifier you replace "such that" by implication and in existential quantifier you replace it by conjuction. So "for every x such that x is a man, x is mortal" is written with implication and "for every". And "there exists x such that it is trigonometric, which is periodic" is written with conjunction (you meant that, disjuction is not used).
I can explain it by describing universal quantifier by the existential one. $\forall \varphi(x)$ is equivalent to $\neg \exists \neg \varphi(x)$. Now, when you add the "such that" clause $\psi(x)$, you add it after that existential quantifier but you do not negate it: $\neg \exists \psi(x)\wedge\neg \varphi(x)$ is equivalent to $\neg \exists \neg(\psi(x)\rightarrow\varphi(x))$, so that means $\forall \psi(x)\rightarrow\varphi(x)$.
You have to consider the interplay of quantifiers and propositional conncetives.
When we write : "Some trigonometric functions are periodic" we want to assert that "There is at least one function that is trig and periodic".
This is represented as :
$\exists x \ (Trigonometric(x) \land Periodic(x))$.
If instead we write : $\exists x \ (Trigonometric(x) \to Periodic(x))$ what happens with a function $f$ whatever that is not trig?
By truth table for conditional ($\to$) we have that $f$ satisfy the formula: $Trigonometric(x) \to Periodic(x)$ (because F → F is T) and thus the said function is a "witness" that satisfy the sentence : "There is at least one..." and this is not what we want.
With "All men are mortal", if we represent it with :
$\forall x \ (Man (x) \land Mortal (x))$,
that reads : "For all objects in the "universe", they are man and mortal", what we get is a false statement in a domain that includes also "objects" that are not men.
My dog Fido is not a man and we have that: $Man(Fido) \land Mortal(Fido)$ is false (F ∧ T is F) and thus we cannot assert that "For all objects...", and again this is not what we want.