Ok, so this is a confusing one. I'm not sure what my teacher is looking for. The problem is:
Plot any number $-\pi < y< \pi$ and pick a small number $\delta > 0$ so that the whole interval $I = (y-\delta, y+\delta) ⊆ T$ with $T=(-\pi, \pi)$.
Now define a function $f(x) =\cos(x-y)-\cos(\delta)$ and plot it on the interval $T$. Use the fact that on $T$ the cosine function takes on no value more than twice to show that:
a)$f(x)=0, x=y-\delta, y+\delta$
b)$f(x)>0$ on $ I$
c)$f(x)<0$ otherwise
I'm guessing the solution is some kind of graph, but I can't seem to find a clear graph of this function. Help! Please!
a) Recall that $\cos x=\cos (-x)$ for all $x$
b+c) In a) you've found two points at which $f=0$, so by the hint $f\neq 0$ everywhere else. So you just have to find out whether $f(x)>0$ for $x\in (y-\delta,y+\delta)$ or outside.