How to show that this real function is not periodic? 
How can one prove that $$\cos\left(\frac{\pi}{2} t \right)+\cos\left(t \right)$$ is not periodic?

This question is motivated by the harmonic spectral representation of time series. Indeed, it is easy to show that a path of a time series given by $$ \cos(\lambda_1 t) + \cos(\lambda_2 t)$$ is periodic if $\frac{\lambda_1}{\lambda_2} \in \mathbb{Q}$. In the above example this is not the case since $\frac{\lambda_1}{\lambda_2} = \pi/2.$
So the case above could serve as an example for the statement: if $\frac{\lambda_1}{\lambda_2} \not\in \mathbb{Q}$ then the path of a harmonic time series is in general not periodic.
 A: The general case solves similarly. 
Suppose that $ f(t) = \cos(\lambda_1 t) + \cos(\lambda_2 t)$ is periodic = $f(t + T)$
Then $f(0) = \cos(0) + \cos(0) = 2 = \cos(\lambda_1 T) + \cos(\lambda_2 T)$
So, $\lambda_1 T = 2\pi n$ and $\lambda_2 T = 2\pi m$ for integers n and m 
Therefore  $\frac{\lambda_1}{\lambda_2}  = n/m$ i.e. $\frac{\lambda_1}{\lambda_2}  \in \mathbb{Q}$
(assuming $\lambda_2, T$ are not zero).
A more general case is  $ f(t) = c_1\cos(\lambda_1 t) + c_2\cos(\lambda_2 t)$ 
Assume firstly that $c_1$ and $c_2$ are the same sign. Then similar to the above $f(0) = c_1 \cos(0) + c_2 \cos(0) = c_1 + c_2 = c_1 \cos(\lambda_1 T) + c_2 \cos(\lambda_2 T)$
Since $c_1$ and $c_2$ are the same sign this still requries that $\lambda_1 T = 2\pi n$ and $\lambda_2 T = 2\pi m$ for integers n and m and the result follows.
If they are different signs, then my answer posted earlier was wrong,
And, an even more general case is  $ f(t) = c_1\cos(\lambda_1 t +\alpha_1) + c_2\cos(\lambda_2 t + \alpha_2)$ 
Here, use the expansion of $\cos (a + b) = \cos(a)\cos(b) + \sin(a) \sin(b)$.
The function $f$ is now of the form $ f(t) = d_1\cos(\lambda_1 t) + d_2\cos(\lambda_2 t) + d_3\sin(\lambda_1 t) + d_4\sin(\lambda_2 t) $ . 
My earlier conclusion of this part of the proof was wrong, so this is part of the question is still open.
A: The function 
$$f(t)=\cos\left(\frac{\pi}{2} t \right)+\cos\left(t \right)$$
is even; so $f(t+T)=f(t)  \Leftrightarrow f(t-T)=f(t),~~ \forall t$.
Adding the equations
$$f(t+T)=f(t), $$
$$f(t-T)=f(t), $$
we arrive at 
$$\cos t \cos T + \cos\left(\frac{\pi}{2} t \right)\cos\left(\frac{\pi}{2} T \right)=
\cos t+ \cos\left(\frac{\pi}{2} t \right),~~ \forall t. $$
Then, for $t=0$ we have
$$ \cos T + \cos\left(\frac{\pi}{2} T \right)= 2 $$
which implies $T=2m\pi$ and $T=4n$, for $n,m\in\mathbb Z$. In summary, there exists no solution to the periodicity problem ($T=0$ is no period in our conventions).
