You call it Time period I call it a Question

Let f(x)=|sin x|+|cos x|. Then,

f(x+π/2)=|sin(x+π/2)|+|cos(x+π/2)|=|cos x|+|-sin x|=f(x).

Therefore, if it can be shown that T=π/2 is the least positive number satisfying f(x+T)=f(x), then the period of f is π/2. Suppose that f(x+T)=f(x), with 0<T<π/2. Then,

|sin(x+T)|+|cos(x+T)|=|sin x|+|cos x|.                                                                                                                Eq.(1)

Squaring both sides in Eq.(1) and simplifying, using cos²θ+sin²θ=1 and sin(2θ)=2sinθcosθ, gives

|sin(2x+2T)|=|sin(2x)|.                                                                                                                                       Eq.(2)

Squaring both sides in Eq.(2) gives

sin²(2x+2T)=sin²(2x),                                                                                                                                        Eq.(3)

or

[sin(2x+2T)-sin(2x)][sin(2x+2T)+sin(2x)]=0,                                                                                                     Eq.(4)

after factoring the difference of two squares. Setting the first factor in Eq.(4) equal to zero, gives

sin(2x+2T)=sin(2x),

which means that 2T is a multiple of the period, π, of the function x↦sin(2x). Therefore, 2T=nπ for some positive integer n. But this contradicts the assumption that 0<T<π/2. Setting the second factor in Eq.(4) equalto zero, gives

sin(2x+2T)+sin(2x)=0.                                                                                                                                       Eq.(5)

Using the trigonometric identity

sinα+sinβ=2sin(((α+β)/2))cos(((α-β)/2))

in Eq.(5) gives

sin(2x+T)cosT=0.                                                                                                                                              Eq.(6)

Therefore,

cosT=0 or sin(2x+T)=0.                                                                                                                                     Eq.(7)

The first choice in Eq.(7) is not possible since cosT is positive when 0<T<π/2. The second condition in Eq.(7) implies that T is a multiple of the period, π, of the function x↦sin(2x), giving

T=nπ, for some integer n,

again contradicting the assumption that 0<T<π/2. Therefore, π/2 is the smallest positive number satisfying f(x+T)=f(x). Therefore the period of f is π/2. This was typed using Scientific Workplace and copied and pasted here. There seems to be some incompatibility issue in the way the two environments handle LATEX.

Since sine and cosine have both periods $2 \pi$ , that case is trivial.

To prove period $\pi$ we can use identities $(\forall x \in \mathbb{R}) \sin(x)=-\sin(x+\pi)$ and $\cos(x)=-\cos(x+\pi)$ as follows: $\begin{aligned} f(t + \pi) &= |\sin (t+\pi)| + |\cos(t+\pi)|\\ &= |- \sin (t)| + |- \cos (t)|\\ &= |\sin (t)| + |\cos (t)|\\ &= f(t) \end{aligned}$

To prove the $\pi/2$ case we use angle addition identities and known values of sine and cosine for $\pi/2$ : $\begin{aligned} f\left(t+\frac{\pi}{2}\right)&= \left | \sin\left( t+\frac{\pi}{2}\right ) \right|+\left|\cos\left(t+\frac{\pi}{2}\right)\right|\\ &= \left|\cos\left(\frac{\pi}{2}\right) \sin (t) + \sin \left(\frac{\pi}{2}\right) \cos (t) \right| + \left|\cos \left(\frac{\pi}{2}\right) \cos (t) - \sin \left(\frac{\pi}{2}\right) \sin (t) \right|\\ &= | 0 \sin (t) + 1 \cos (t) | + | 0 \cos (t) - 1 \sin (t) |\\ &= |\sin (t)|+|\cos (t)|\\ &= f(t) \end{aligned}$

$\pi/4$ answer can be proven wrong by counterexample e. g. knowing that $\sin(\pi/4) = \cos(\pi/4) = \sqrt{2}/2$ we have

$\begin{aligned} f\left(0+\frac{\pi}{4}\right) &= \left|\sin\left(0+\frac{\pi}{4}\right)\right| + \left|\cos\left(0+\frac{\pi}{4}\right)\right|\\ &= \frac{2\sqrt{2}}{2} = \sqrt{2}\\ &\neq\\ f(0) &= |\sin(0)| + |\cos(0)| = 1 \end{aligned}$ Hence it can't be true that $(\forall t \in \mathbb{R}) f(t+\pi/4)=f(t)$ and the answer is $\pi/2$ .

$\sin t$ and $\cos t$ have periods of $2\pi$

$|\sin t|$ and $|\cos t|$ have periods of $\pi$

But their sum has a period of $\pi /2.$ Can someone explain why?