Question: We know that $\varphi(t) = \cos{t}$ is a characteristic function of Rademacher random variable. Is the function $\phi(t) = \cos{t} \cdot 1\_{|t| \le \pi /2}$ still a characteristic function of some random variable?

The only thing we have to check is if $\phi$ is positively-defined. I tried looking for some examples contradicting it, but with no success, so I suspect it is a characteristic function. When I write the condition of positive-definiteness of $\phi$, it looks very similar to the condition for $\varphi$, but with more zeros, but I cannot deduce anything out of it.

Answer: 
If your given function $\phi (t) = {\Bbb E}(e^{ i t X})$ is the characteristic function of a random variable $X$ whose probability density is $f(x)dx$, then $ \phi(t)$ is the Fourier transform of $f(x)$ and the extremely useful **Fourier inversion formula** allows us to recover the function $f(x)$ by the general identity $$f(x) = \frac{1}{2\pi} \int\_{t=-\infty}^{t=\infty} e^{- i t x} \phi(t) dt$$ (See footnote regarding sign conventions.)

In your example this simplifies to $f(x) =\frac{1}{2\pi} \int\_{t=-\pi/2}^{t=\pi/2} \ (\cos t ) \ e^{-i t x} dt$ which turns out to be $$f(x)=\frac{ \cos(\pi x/2)}{ \pi ( 1- x^2)}$$

Plotting the graph of $f(x)$, we see that it fails to be always non-negative; thus $f(x)$ is NOT a probability density.
[![enter image description here](https://i.stack.imgur.com/wVi8k.png)](https://i.stack.imgur.com/wVi8k.png)

P.S. Wikipedia and many other sources use a slightly different convention for the Fourier transform (engineer's version) that shuffles the constant $2\pi$ into other places. The version used above is what is used in statistics and in theoretical physics.
See e.g equations 3 and 4 in reference below.

[Physicist's Fourier formulas](https://physics.stackexchange.com/questions/308234/fourier-transform-standard-practice-for-physics)

P.P.S. In response to a comment that raises a subtle issue I would add one clarification.

There is no a priori assumption that $X$ has a density. Since $\phi$ has compact support there is no doubt that it is the Fourier transform of a smooth function $f(x)$. And the uniqueness theorem for Fourier transforms of generalized functions implies that no finite measure $X$ could exist that has the same characteristic function as this $f(x)$. Thus ultimately the pivotal question is whether or not this function $f(x)$ is non-nonnegative.

The other characteristic function mentioned, $\cos t$ did not decay at $\infty$ so by the contrapositive of the Riemann-Lebesgue Lemma it could not be the characteristic function of any absolutely integrable function. Instead it is of course the Fourier transform of a pair of Dirac delta functions (point masses) which are of course generalized functions.
Suppose $\phi$ is the Fourier transform of a finite measure $\mu$ on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$. [By Lévy inversion](https://www.wolframalpha.com/input?i=integrate+%28e%5E%7Bi5x%7D-e%5E%7Bi3x%7D%29%2F%28ix%29*cos%28x%29+from+-pi%2F2+to+pi%2F2)
$$\begin{aligned}\frac{\mu(3)+\mu(5)}{2}+\mu((3,5))&=\int\_{[-\pi/2,\pi/2]}\frac{e^{i5\xi}-e^{i3\xi}}{i\xi}\cos(\xi)d\xi\approx -0.18\end{aligned}$$
But $\mu(B)\geq 0$ for all Borel $B$, so $\phi$ is not the CF of a finite measure $\mu$.