Why does the plot of $f(x)=|cos x|-|sin x|$ look almost piecewise linear?

• I Quadrant$rightarrow xinleft(0,frac{pi}2right], f(x) = lvert cos x rvert - lvert sin x rvert=cos x-sin x=-sqrt2sinleft(x-frac{pi}4right)$
• II Quadrant$rightarrow xinleft(frac{pi}2,piright], f(x) = lvert cos x rvert - lvert sin x rvert=-cos x-sin x=-sqrt2sinleft(x+frac{pi}4right)$
• IV Quadrant$rightarrow xinleft(-frac{pi}2,0right], f(x) = lvert cos x rvert - lvert sin x rvert=cos x+sin x=sqrt2sinleft(x+frac{pi}4right)$
• III Quadrant$rightarrow xinleft(-pi,-frac{pi}2right], f(x) = lvert cos x rvert - lvert sin x rvert=-cos x+sin x=sqrt2sinleft(x-frac{pi}4right)$

Draw individual graphs in individual domains and you should get the desired graph. Also, it looks linear just because of the scale you have chosen. Actually, it is as sinusoidal as sine is in the intervals of $frac{pi}2$.

Of course, the function must have period at most $2 pi$. Over $[0,2 pi]$, we have $$ f(x) = |cos(x)| - |sin(x)| = begin{cases} cos(x) - sin(x) & 0 leq x < pi/2\ -cos(x) - sin(x) & pi/2 leq x < pi\ -cos(x) + sin(x) & pi leq x < 3pi/2\ cos(x) - sin(x) & pi leq x leq 2 pi. end{cases} $$ With the sum to product identities, we can rewrite this as $$ f(x) = begin{cases} sqrt{2}sin(pi/4 - x) & 0 leq x < pi/2\ sqrt{2}sin(x - 3 pi/4) & pi/2 leq x < pi\ -sqrt{2}sin(pi/4 - x) & pi leq x < 3pi/2\ -sqrt{2}sin(x - 3 pi/4) & pi leq x leq 2 pi end{cases} implies\ f(x) = begin{cases} sqrt{2}sin(pi/4 - x) & 0 leq x < pi/2\ sqrt{2}sin(x - 3 pi/4) & pi/2 leq x < pi\ sqrt{2}sin(pi/4 - (x - pi)) & pi leq x < 3pi/2\ sqrt{2}sin((x-pi) - 3 pi/4) & pi leq x leq 2 pi. end{cases} $$ In other words, we see that $f(x)$ in fact has a period of $pi$, and over the single period $[0,pi]$ it can more simply be written in the form $$ f(x) = begin{cases} sqrt{2}sin(pi/4 - x) & 0 leq x < pi/2\ sqrt{2}sin(x - 3 pi/4) & pi/2 leq x leq pi. end{cases} $$ Now, we can conveniently answer the question at hand: why does $f(x)$ look piecewise linear? One answer is that because we have $sin(x) approx x$ when $|x|$ is sufficiently small, $f(x)$ is well approximated by $$ f(x) approx begin{cases} sqrt{2}(pi/4 - x) & 0 leq x < pi/2\ sqrt{2}(x - 3 pi/4) & pi/2 leq x leq pi. end{cases} $$

Here is a plot of the two functions, for comparison.