Show that $\{ \frac 12 (1+\cos (x-m\pi))\}$ forms a partition of unity Exercise.$1$(Munkres- pg.$140$) :  let $f:\mathbb R \to \mathbb R$ be defined by the equation
$$f(x)= \left\{ \begin{array}{lcc}
             \frac{(1+\cos x)}{2} &   \text{for }   -\pi \leq x \leq \pi \\
             \\ 0 &    \text{otherwise}.
             \end{array}
   \right.$$
Then $f$ is of class $C^1$. For each integer  $m \geq 0$, set $\phi_{2m+1}(x)=f(x-m\pi)$. For each integer $m \geq1$, set $\phi_{2m}(x)=f(x+m\pi)$. Then the collection {$\phi_i$} forms a partition of unity on $\mathbb R$. The Support $S_{i}$ of $\phi_{i}$ is a closed interval of the form $[k\pi,(k+2)\pi]$,which is compact,and each point of $\mathbb R$ has neighborhood  that intersects at most three of the sets $S_{i}$. We leave it to you to check that $\sum \phi_{i}(x)=1$. thus {$\phi_{i}$} is a partition of unity on $\mathbb R$.
I did the following: if $\phi_{2m+1}(x)=f(x-m\pi)$ then $-\pi\leq x-m\pi\leq \pi$ imply $\pi(m-1)\leq x\leq \pi(m+1)$, the same way for $\phi_{2m}$ we have $-\pi(m+1)\leq x\leq -\pi(m+1)$. The support is adherent set of point of domain where $\phi_i$ is different of zero so I put
$$\operatorname{supp} \phi_i=\overline{[\pi(m-1),\pi(m+1)]\cup [-\pi(m+1),-\pi(m+1)]},$$
but the book wrote
$$\operatorname{supp} \phi_i=[k \pi,(k+2)\pi].$$
Is it a mistake because it has to be $m$ right?
Besides I need a hint to show the last part. I tried to do with $m=0$ and $m=1$, $\phi_1= \frac{1+\cos x}{2}$ and $\phi_2=\frac{1+\cos(x-\pi)}{2}$ and the sum of $\phi_1$ and $\phi_2$ is equal to $1$. But how to generalize this and how to find that just three set of point have intersection in the support?
 A: For a function $f : X \to \mathbb R$ let $s(f) = \{ x \in X \mid f(x) \ne 0 \}$. Then the support of $f$,  $\text{supp} f$,  is defined as the closure of $s(f)$ in the space $X$. To answer your question it will be easier to work with $s(-)$ than with $\text{supp}$.
For your function $f : \mathbb R \to \mathbb R$ we have $s(f) = (-\pi,\pi)$. Thus $s(\phi_{2m+1}) = ((m-1)\pi,(m+1)\pi)$ and $s(\phi_{2m}) = ((-m-1)\pi,(-m+1)\pi)$ because $\phi_i$ is obtained from $f$ by a shift of the variable $x$.
This shows that $\text{supp}\phi_i$ always has the form $[k\pi,(k+2)\pi]$ for some $k = k(i)$.
To verify that $\Sigma(x) = \sum \phi_i(x) = 1$ for all $x$, note that only the indices $i$ make a contribution $> 0$ for which $x \in s(\phi_i)$. We have  $s(\phi_1) = (-\pi,\pi)$, $s(\phi_2) = (-2\pi,0)$, $s(\phi_3) = (0,2\pi)$, $s(\phi_4) = (-3\pi,-\pi)$, $s(\phi_5) = (\pi,3\pi)$, etc.
Case 1: $x = r\pi$ with $r \ge 0$. We have $r \pi \in s(\phi_i)$ precisely for $i=2r+1$. This is true because for $i = 2m$ with $m \ge 1$ we have $s(\phi_i) \subset (-\infty,0)$ and for $i = 2m+1$ with $m \ge 0$ we have $r\pi \in s(\phi_i) =  ((m-1)\pi,(m+1)\pi)$ iff $r = m$, i.e. $i = 2r+1$.
Thus $\Sigma(r \pi)= \frac{1 + \cos(r \pi - r \pi)}{2} = 1$.
Case 2: $x = r\pi$ with $r<  0$. Then $x  \in s(\phi_i)$ precisely for $i=-2r$.  This is true because for $i = 2m+1$ with $m \ge 0$ we have $s(\phi_i) \subset (-\pi,\infty)$ and for $i = 2m$ with $m \ge 1$ we have $r\pi \in s(\phi_i) =  ((-m-1)\pi,(-m+1)\pi)$ iff $r = -m$, i.e. $i = -2r$.
Thus $\Sigma(r \pi)= \frac{1 + \cos(r\pi + (-r)\pi)}{2} = 1$.
For the next two case recall that $\cos(y\pm\pi) = -\cos(y)$.
Case 3: $x \in (r\pi, (r+1)\pi)$ with $r \ge 0$. Then $x  \in s(\phi_i)$ precisely for $i=2r+1$ and $r = 2(r+1)+1$. This is true because for $i = 2m$ with $m \ge 1$ we have $s(\phi_i) \subset (-\infty,0)$ and for $i = 2m+1$ with $m \ge 0$ we have
$$(r\pi, (r+1)\pi) \cap s(\phi_i) =  (r\pi, (r+1)\pi) \cap ((m-1)\pi,(m+1)\pi) = \begin{cases} \emptyset & m \ne r, r+1 \\  (r\pi, (r+1)\pi) & m = r, r+1\end{cases}$$
Thus with $y = x + \pi r$ we have $\Sigma(x)= \phi_r(x) + \phi_{r+1}(x) = f(y) + f(y +\pi) = \frac{2 + \cos(y) + \cos(y + \pi)}{2} = 1$.
Case 4: $x \in (r\pi, (r+1)\pi)$ with $r < 0$. This is similar as case 3.
