Induction Proof of trig inequality $\sum_{k=0}^n |\cos k| \ge \frac n2$ This is for a course, I don't want the answer just a prod in the right direction!
I've got a problem that states
let n be an integer such that $$n\gt0$$ $$\text{Prove: }\sum_{k=0}^n |\cos k| \ge \frac n2$$
I'm using induction to prove this. First I showed my base case:
When $P(0)$ $\sum_{k=0}^0 |\cos o| \ge \frac 02$
Which simplifies down into $|1|\ge 0$ which holds true. So I moved to the induction step.
Assume: $P(n)$ is true such that $\sum_{k=0}^n |\cos k| \ge \frac n2$
WTS: P(n+1) is true such that $\sum_{k=0}^{n+1} |\cos k| \ge \frac n2$
So I broke the summation up into two parts
$\sum_{k=0}^{n+1} |\cos k|=\sum_{k=0}^n |\cos k|+ \cos(n+1)$ Which I can substitute back in and get
$\sum_{k=0}^n |\cos k|+ \cos(n+1) \ge \frac {n+1}2$
I know, well I think, that this is where I use my Induction Hypothesis $\sum_{k=0}^n |\cos k| \ge \frac n2$
But I don't now how to show my next step.
Can I just substitue $\frac n2$ in for $\sum_{k=0}^n |\cos k|$ and continue to solve? or do I have to show more steps in the middle? or am I completely lost? If anyone can prod me into the right direction it would be much appreciated! Thanks for your help!
First Edit
@Bungo
This is the way I've been doing it.
Is my logic wrong? Do I need to prove the lower bound? Or is my logic okay?
So I broke the sum down and re wrote it as $\sum_{k=0}^n + |\cos n+1| \ge \frac {n+1}2$
And then substituted from the induction hypothesis and wrote it as $\frac n2 + |\cos n+1| \ge \frac {n+1}2$
Then I simplified and got $|\cos n+1| \ge \frac 12$
So now I'm trying to prove that statement as true. Is this okay? Or should I go back and try with the triplets as you suggested? Thanks!
 A: The statement
$$\sum_{k=0}^{n}|\cos(k)| \geq \frac{n}{2}$$
appears to be true, but it is not in general true that
$$\frac{n}{2} + |\cos(n+1)| \geq \frac{n+1}{2}$$
(e.g., this is false for $n=1$). So I don't think it will be useful to decompose the sum as $$\sum_{k=0}^{n+1} |\cos(k)| = |\cos(n+1)| + \sum_{k=0}^n |\cos(k)|$$
because we don't have a lower bound for $|\cos(n+1)|$ other than zero.
Instead, one might look at pairs of the form $|\cos(n)| + |\cos(n+1)|$ and try to find a useful lower bound for these. Plotting $|\cos(x)| + |\cos(x+1)|$ at Wolfram Alpha suggests that $|\cos(x)| + |\cos(x+1)| > 0.8$ for all $x$, which is still not good enough: we need a lower bound of $1$ if we add in pairs.
So let's look at triplets: $|\cos(n-1)| + |\cos(n)| + |\cos(n+1)|$. Wolfram Alpha's plot shows that the function $|\cos(x-1)| + |\cos(x)| + |\cos(x+1)|$ stays above $1.5$, so if you can prove that lower bound, breaking the sum into groups of triplets should suffice. I think proving the lower bound will be grungy, though. Hopefully someone can propose a nicer solution.
A: Usually in Induction we do the following:

*

*Show that the relation is true for the case $k=1$, i.e., prove that $\mathfrak{R}(1)$ is true, usually by putting values.

*Assume truth of the relation for a general value of $k$, i.e., assume $\mathfrak{R}(k)$ is true.

*Show that by assuming truth for a value of $k$ proves that the relation is also true for $(k+1)$, usually by showing that $\mathfrak{R}(k+1)=f(\mathfrak{R}(k))$* and then putting the expression of $\mathfrak{R}(k)$ into this and then establishing truth of $\mathfrak{R}(k+1)$.

*If this is done, you have proved the theorem/identity by induction.


[*or in simple words, express $\mathfrak{R}(k+1)$ in terms of $\mathfrak{R}(k)$, let this correlation be denoted by $f$, usually but not always you get $\mathfrak{R}(k+1)=\mathfrak{R}(k)+\lambda$ or $f=\mathfrak{R}(k)+\lambda$]

Trying to use induction in this case:

*

*$P(1)$ is true.

*Assuming $P(n)$ is true such that $\sum_{k=0}^n |\cos k| \ge \frac n2$

*Now trying to prove that $P(n+1)$ is true using $P(n)$, or:
$$P(n+1)=P(n)+|\cos(n+1)|\ge\frac n2+|\cos(n+1)|\ge^*\frac{n+1}2$$
The $^*$ relation is not always true, so it's not possible to prove by this simple induction, you have to evoke something more complex.


If you are not bound to use induction you may use the result:
$$\cos\alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+\cdots+\cos\left(\alpha+(n-1)\beta\right)\\=\sum_{k=0}^{n-1}\cos\left(\alpha+k\beta\right)=\frac{\cos\left(\alpha+\frac{n-1}2\beta\right)\sin\left(\frac{n\beta}2\right)}{\sin\left(\frac{\beta}2\right)}$$
