# Prove |cos(x−1)|+|cos(x)|+|cos(x+1)|≥3/2

I'm working on an induction proof, but I keep coming up against a brick wall.

While working through the induction proof process I keep ending up with $$|\cos(m)|\ge\frac12$$ ,but clearly this isn't true for all m. From another post someone gave me the idea to look at triples and to consider $$|\cos(x−1)|+|\cos(x)|+|\cos(x+1)|$$ to try and find a lower bound. If you plug the above equation into Wolfram Alpha it tells you that $$|\cos(x−1)|+|\cos(x)|+|\cos(x+1)|\ge\frac32$$ Which would prove my induction proof. I just don't know how to go about proving that it does truly exceed $\frac32$

If it helps the original statement that I was trying to prove is $$\sum_{k=0}^n|\cos(k)|\ge\frac n2$$

Well I have a proof for you but its not the most elegant.

Assume first that $\cos(x+1)$ and $\cos(x-1)$ have opposite signs, then $$|\cos(x+1)|+|\cos(x-1)|=|\cos(x+1)-\cos(x-1)|=|2\sin (1) \sin x|$$

Now $$|2\sin 1 \sin x+\cos x |\leq |2\sin 1 \sin x|+|\cos x|$$

Now a bit of calculus, the minimum of $a\sin x+\cos x$ occurs when $\tan x =a$, so its minimum value is $$a\sin x+\cos x=\cos x(a\tan x+1)=\cos x (\tan^2 x+1) =\sec x=\sqrt{a^2+1}$$

Now $\frac{3}{2}\leq \sqrt{a^2+1}$ holds if $\frac{\sqrt{5}}{2} \leq a$ and since $a=2\sin 1$ we have to see that $\frac{\sqrt{5}}{4} \leq \sin 1$. But $\frac{\pi}{4} < 1$ so we are reduced to showing $\frac{\sqrt{5}}{4} \leq \frac{1}{\sqrt{2}}$ which is easy.

Now consider what happens if $\cos(x+1)$ and $\cos(x-1)$ have equal signs. Just to fix ideas lets assume they are positive. Then for an appropriate $n$, $$-(\frac{\pi}{2}-1) < x+2\pi n < \frac{\pi}{2}-1$$ so in particular

$$\sin 1=\cos (\frac{\pi}{2}-1)\leq \cos x$$. Now in this case we have

$$|\cos(x)|+|\cos(x+1)|+|\cos(x-1)|=|\cos(x)+\cos(x+1)+\cos(x-1)|=|\cos(x)(2\cos 1 +1)|$$

Thus with the above inequality we have to show

$$\frac{3}{2}\leq |\sin 1(2\cos 1 +1)|.$$

Now $\sin 1(2\cos 1 +1)=\sin 2+\sin 1$

using $2 < \frac{2\pi}{3}$ and $\frac{\pi}{4}<1$ we have

$$\frac{\sqrt{3}}{2}\leq \sin 2$$ $$\frac{\sqrt{2}}{2}\leq \sin 1$$

So we have $$\frac{3}{2}\leq \frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}$$ which is easily checked.

• What if they limit you to x being a positive integer? Would the proof still hold? – Fmonkey2001 Sep 5 '14 at 1:07
• I dont understand doesnt the proof hold for all $x$ ? – Rene Schipperus Sep 5 '14 at 1:17
• I believe so, but the proof said to assume that x is a non negative integer. So I didn't know if I could use $\frac{\pi}2$ in my proof or not. – Fmonkey2001 Sep 5 '14 at 12:35
• I didnt say that. – Rene Schipperus Sep 5 '14 at 14:13