Period of a function? I am trying to find out the period of a function but this function is giving me a different answer from what I expected:
\begin{equation*}
f(x) = |\sin x| + |\cos x| .
\end{equation*}
I know that to find the  period of $\sin$ and $\cos$ we use the formula $2\pi/ |n|$ , where $n$ is the co-efficient of $x$ . Since, this question contains absolute value of sin n cos , so there respective periods will get cut in half . So , according to me the answer of this question should be $\pi$ but it is not. It's answer is $\pi/2$. Please explain. Thanks !
 A: Indeed the period of $|\sin x|$ is $\pi$, and the period of $|\cos x|$ is also $\pi$. This means that, for every $x$, you have
$$f(x+\pi)=|\sin(x+\pi)| + |\cos(x+\pi)| = |\sin x| + |\cos x| = f(x)$$
so it would seem like $\pi$ is the period of $f$, right?
Wrong. The period of a function is defined as the smallest constant for which $f(x+c)=f(x)$ for all values of $x$. That is why the function $\sin x$ has a period of $2\pi$ and not, say, $26\pi$, even though we know that $\sin(x+26\pi)=\sin x$ for all values of $x$.
This means that you still have to find the period of $f$. You only know that the period will be some fraction of $\pi$ (because $\pi$ is a "candidate period"), but you did not exclude the possibility that the period is $\frac\pi n$ for some $n$.
In order to see what the actual period of $f$ is, I advise you to plot your function on $[0,\pi]$. First, plot it on $[0,\frac\pi 2]$, where $\sin x, \cos x\geq 0$, then on $\frac\pi2, \pi]$, where $\cos x \leq 0$
A: If the period of a function, $f(x)$, is $\dfrac{2\pi}{k}$ for some integer, $k$, then the value of $k$ will readily be apparent when you graph the polar equation $r = f(\theta)$. Below is a plot of 
$$r = |\cos \theta| + |\sin \theta|.$$

It should be apparent that the period is $\dfrac{2\pi}{4} = \dfrac{\pi}{2}$
Actually, this doesn't prove that the period is $\dfrac{\pi}{2}$. This is a start.
\begin{align}
   f\left( x + \dfrac{\pi}{2} \right)
   &=  \left|\cos\left( x + \dfrac{\pi}{2} \right)\right| 
     + \left|\sin\left( x + \dfrac{\pi}{2} \right)\right|
\\
   &=  \left|
          \cos(x) \cos\left(\dfrac{\pi}{2}\right) -
          \sin(x) \sin\left(\dfrac{\pi}{2}\right)
       \right| 
     + \left|
          \sin(x) \cos\left(\dfrac{\pi}{2}\right) +
          \cos(x) \sin\left(\dfrac{\pi}{2}\right)
       \right|
\\
   &=  |-\sin(x)| + |\cos(x)|
\\
   &= f(x)
\end{align}
A: The maximum value of the function is $√2$. It is obtained at $π/4, 3π/4.....$. 
