Prove that $257 + 32\cos 4t > 256 \cos^2t$

Question

Recently I found a problem in math SE(see here) that ask on how many roots a polinomial has inside a ring and lead me to the following inequality $$257 + 32\cos 4t > 256 \cos^2 t$$ for all real $t$.

I've got a proof but I have doubt with it.

My proof Let $f(t) = 257 + 32 \cos 4t - 256\cos^2 t$. Note that $f$ is periodic. We will find the stationary points of $f$. From $f'(t) = 0$ we have $64 \sin 4t - 256 \sin 2t = 0$. Therefore $\sin 2t(\cos2t - 2) = 0$. Thus, $\sin 2t = 0$. From this we have $$f(t_0) = 257 -256\cos^2 2t_0 > 0$$ For all stationary point of $f$.

Is this sufficient to prove the inequality?

I'm also looking another proof of this inequality without usage of derivative.

Thanks

Answer 1

Your method is correct, but has some errors.

You are considering $f(t) = 257 + 32 \cos 4t - 256\cos^2 t$ and want to prove $f(t_0)>0$, where $t_0$ is a critical point (for both max and min stationary points).

So: $$f'(t)=\color{red}{-128}\sin4t\color{red}+256\sin 2t=0 \Rightarrow \sin 2t(\cos 2t-1)=0 \Rightarrow \sin 2t=0 \ \ \text{or} \ \ \cos2t=1 \Rightarrow \\ 2t=\pi k \ \ \text{or} \ \ 2t=2\pi n \Rightarrow 2t=\pi k \Rightarrow t_0=\frac{\pi k}{2}.$$ So: $$f(t_0)=257+32\cos (2\pi k)-256\cos ^2\left(\frac{\pi k}{2}\right)=\\ =256\left[1-\cos ^2\left(\frac{\pi k}{2}\right)\right]+33\ge 33>0. $$

Answer 2

Writing $\cos2t=c,$

using $\cos2x=2\cos^2x-1,$

$$257+32(2c^2-1)-256\cdot\dfrac{1+c}2=64c^2-128c+97=64(c-1)^2+97-64\ge33$$ for real $c$

Answer 3

Your proof looks OK to me on a quick viewing.

For the other part of the proof (the "no derivatives"), you could write $$ \cos 4t = 2 \cos^2 2t - 1 $$ and then you can rewrite $$ \cos 2t = \cos^2 t - 1 $$ so that your original formula ends up as a quartic in $\cos t$...but one involving only even powers. So you can let $y = \cos^2 t$ and get a quadratic in $y$, which you can then analyze by algebra 1 methods.

Answer 4

Note that\begin{align}32\cos(4t)&=32\bigl(2\cos^2(2t)-1\bigr)\\&=32\bigl(2\bigl(2\cos^2(t)-1\bigr)^2-1\bigr)\\&=32\bigl(8\cos^2(t)-8\cos(t)+1\bigr)\\&=256\cos^2(t)-256\cos(t)+32\\&\geqslant256\cos^2(t)-256+32\\&>256\cos^2(t)-257.\end{align}Your proof looks correct to me.

Answer 5

Using $\cos(2x)=2\cos^2(x)-1$ twice we have $257+32\cos(4t)-256\cos^2(t)=257+32(2(2\cos^2(t)-1)^2-1)-256\cos^2(t)$

Simplification leads to

$257+256\cos^4(t)-512\cos^2(t)=257+256(\cos^2(t))^2-512\cos^2(t)=1+(16-16\cos^2(t))^2\geq 1$

The function is periodic and max value occurs at $t=n\pi$ thus we see that the inequality $257+32\cos(4t)-256\cos^2(t)>0$ is true.