$f(x)=\sin(\frac{π}4(x-[x]))$ if $[x]$ is odd, $x≥0$, and $\cos(\fracπ4(1-x+[x])$ if $[x]$ is even and $x≥0$. 
QUESTION: Consider the function $f(x)=\sin(\frac{π}4(x-[x]))$ if $[x]$ is odd, $x≥0$, and $\cos(\fracπ4(1-x+[x])$ if $[x]$ is even and $x≥0$. Where, $[x]$ denotes the greatest integer function less than or equal to $x$.
$i)$ Sketch the graph of the function on a plain paper.
$ii)$ Determine the points of discontinuity of $f$ and the points where it is not differentiable.

MY APPROACH: I changed $(x-[x])$ to $\{x\}$ where $\{x\}$ denotes the fractional part of $x$.
$\because$ $\{x\}$ belongs to $(0,1)$ $\therefore$ $\sin(\fracπ4\{x\})$ belongs to $(0,\frac1{√2})$ when $[x]$ is odd. Likewise I did for the cosine part. But this does not seem to be of much use.
Can someone help me out?
Thank you.
 A: The function of interest is $$f(x) =\begin{cases} \sin\left(\frac{\pi}{4}f \right), \  \space \text{if} \ 2n+1 \le x\lt 2n+2  \hspace{1.5 cm} n\in \mathbb N\cup \{0\} \\  \sin \left( \frac{\pi}{4}(1+f) \right), \ \space \text{if} \ 2n\le x\lt2n+1 \end{cases} $$ where $f$ denotes the fractional part of $x$.
Notice that determining the graph for $0\le x \le 2$ will suffice, as the function only depends on $f$.
In $[0,1)$, $f(x)$ increases from $\frac{1}{\sqrt 2}$ to $1$. In $[1,2)$, $f(x)$ increases from $0$ to $\frac{1}{\sqrt 2}$. We have a point of discontinuity at $x=1$, and consequently at all odd integers. 
The only thing that remains to do is to check whether $f(x)$ is differentiable at $x=2$. Note the left derivative is $$\lim_{x\to 2^-} \frac{f(x)-f(2)}{x-2} \\ =\lim_{x\to 2^-}\frac{\sin\left(\frac{\pi}{4}(x-1)\right)-\frac{1}{\sqrt 2}}{x-2} \\ =\frac{\pi}{4\sqrt 2}$$ 
The right derivative, similarly is $\frac{\pi}{4\sqrt 2}$. Hence, $f(x)$ is differentiable at $2$ and consequently at all even numbers. Now you should be able to draw out the graph.
