(1) Plot the graph of the function $y=f(x)$ where $f(x)=\sin x+|\sin x|$. Differentiate the function and say where the derivative is defined and where it is not defined.

I observe that $\sin x$ takes positive values between $2k\pi$ and $(2k+1)\pi$ (for integer $k$), that is in the first two quadrants, and $\sin x$ takes negative values between $(2k+1)\pi$ and $(2k+2)\pi$, that is in the third and fourth quadrants. So

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{f(x)}& =2\sin x\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\mathrm{for}\hspace{0.5em}2k\pi \le x\le (2k+1)\pi \\ \multicolumn{1}{c}{}& =0\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\mathrm{for}\hspace{0.5em}(2k+1)\pi \le x\le (2k+2)\pi .\end{array}}$

Below is represented the graph of y = f(x):

(2) Now I express the function $f(x)=\sin x+\cos x$ in the form $f(x)=A\sin (x+\alpha )$, find $A$ and $\alpha$ and plot the graph of this function. Similarly I express the function $g(x)=\sin x-\cos x$ in the form $g(x)=B\sin (x+\beta )$ where $-\pi /2<\beta <\pi /2$, and plot its graph on the same axes.

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{f(x)}& =\sin x+\cos x=A\sin (x+\alpha )\\ \multicolumn{1}{c}{}& =A\sin x\cos \alpha +A\cos x\sin \alpha \end{array}}$

As this formula must be valid for any x, I obtain:

${\displaystyle A\sin \alpha =1\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\mathrm{and}\hspace{0.5em}A\cos \alpha =1}$

and hence $\tan \alpha =1$, $\alpha =\pi /4$ and $A=\pm \sqrt{2}$.

Comment: If $A$ has the meaning of an amplitude, $A$ is positive, and only the positive solution must be kept. This type of problem is typical for the composition of oscillations.

Hence the graph of this function is a sine graph with a phase shift of $\pi /4$, that is $f(x)=0$ when $x=k\pi -\pi /4$, it takes the value 1 when $x=2k\pi$ and $x=2k\pi +\pi /2$ and the value -1 when $x=(2k+1)\pi$ and $x=2k\pi -\pi /2$, has maximum values $(2k\pi +\pi /4,\sqrt{2})$ and minimum values $((2k+1)\pi +\pi /4,-\sqrt{2})$

In a similar manner I write $g(x)=\sin x-\cos x$:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{f(x)}& =\sin x-\cos x=B\sin (x+\beta )\\ \multicolumn{1}{c}{}& =B\sin x\cos \beta +B\cos x\sin \beta \end{array}}$

So I have

${\displaystyle B\sin \beta =-1\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\mathrm{and}\hspace{0.5em}B\cos \beta =1}$

and hence $\tan \beta =-1$, $\beta =-\pi /4$ and $B=\sqrt{2}$. Hence the graph of this function is a sine graph with a phase shift of $-\pi /4$, that is $f(x)=0$ when $x=k\pi +\pi /4$, it takes the value 1 when $x=k\pi$ and $x=2k\pi +\pi /2$, has maximum values $(2k+1)\pi -\pi /4,\sqrt{2})$ and minimum values $(2k\pi -\pi /4,-\sqrt{2})$.

(3) Now, I calculate the function $f(x)=\sin x+|\cos x|$ and plot the graph.

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{f(x)}& =\sin x+\cos x=\sqrt{2}\sin (x+\pi /4)\hspace{0.5em}\mathrm{for}\hspace{0.5em}2k\pi -\pi /2\le x\le 2k\pi +\pi /2\\ \multicolumn{1}{c}{}& =\sin x-\cos x=\sqrt{2}\sin (x-\pi /4)\hspace{0.5em}\mathrm{for}\hspace{0.5em}2k\pi +\pi /2\le x\le 2k\pi +3\pi /2.\end{array}}$

The derivative is not defined at $x=k\pi +\pi /2$ and for other values of $x$ it is:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{f^{\text{'}}(x)}& =\cos x-\sin x\hspace{0.5em}\mathrm{for}\hspace{0.5em}2k\pi -\pi /2<x<2k\pi +\pi /2\\ \multicolumn{1}{c}{}& =\cos x+\sin x\hspace{0.5em}\mathrm{for}\hspace{0.5em}2k\pi +\pi /2<x<2k\pi +3\pi /2.\end{array}}$