Without Calculator find $$\left\lfloor 2 \cos \left(50^{\circ}\right)+\sqrt{3}\right\rfloor$$ Where $\left \lfloor x \right \rfloor$ represents the floor function.
My Try:
Let $x=2\cos(50^{\circ})+\sqrt{3}$. We have $$\begin{align*} & \cos \left(50^{\circ}\right)<\cos \left(45^{\circ}\right) \\ \Rightarrow \quad & 2 \cos \left(50^{\circ}\right)<\sqrt{2} \\ \Rightarrow \quad & x<\sqrt{3}+\sqrt{2}<3.14 \end{align*}$$
Now I am struggling to prove that $x>3$.
A calculus-based solution:
As you noticed, we only need to prove that $ \cos 50^{\circ}>\frac{3-\sqrt 3}{2}$.
Because of the identity $\cos 3x=4\cos^3x -3\cos x $, we conclude that $\cos 50^{\circ}$ is a root of the equation:
$$\cos 150 ^{\circ}=\frac{-\sqrt 3}{2}=4t^3-3t.$$
However, it is very easy to see that $4t^3-3t+\frac{\sqrt 3}{2}=0$ has three roots; one in the range $(-\infty, -\frac{1}{2} ]$, one in the range $[-\frac{1}{2}, \frac{1}{2} ]$, and the other in the range $[\frac{1}{2}, \infty)$.
Since $\cos 50^{\circ} >\cos 60^{\circ}=\frac{1}{2}$, we must have $\cos 50^{\circ} \in [\frac{1}{2}, \infty) $. Moreover $f(t)=4t^3-3t+\frac{\sqrt 3}{2}$ is increasing for $t>\frac{1}{2}$.
Therefore, we only need to show that $f(\frac{3-\sqrt 3}{2})<0=f(\cos 50^{\circ})$:
$$f(\frac{3-\sqrt 3}{2})=\frac{(3-\sqrt 3)^3}{2}-\frac{3(3-\sqrt 3)}{2}+\frac{\sqrt 3}{2} \\= \frac{((3-\sqrt 3)^2-3)(3-\sqrt 3)}{2}+\frac{\sqrt 3}{2} \\= \frac{(9-6\sqrt 3)(3-\sqrt 3)+\sqrt 3}{2}=\frac{45-26\sqrt 3}{2}<0. $$
Now, we are done because: $45^2=26^2 \times 3-3$.
50° is between 45° and 60° ( $50°=45°+(60°-45°)/3$ ), and in this range, function $cos$ is concave. So $\cos(50°) > \cos 45°+(\cos60°-\cos45°)/3$
$\cos 50° > \sqrt{2}/2 + (1/2- \sqrt{2}/2)/3$.
Approximate $\sqrt{2}$ with $1.414$ and $\sqrt{3}$ with $1.732$, and you are in.
Without derivatives:
Remark: I used this trick in my answer before.
We want to prove that $2\cos \frac{50\pi}{180} + \sqrt 3 > 3$.
Let $u = \cos \frac{50\pi}{180}$ and $v = \frac{3 - \sqrt 3}{2}$. We need to prove that $u > v$.
We have $u \ge \cos \frac{60\pi}{180} = 1/2$ and $v > \frac{3 - 2}{2} = 1/2$. Thus, $4u^2 + 4uv + 4v^2 > 3$.
Thus, it suffices to prove that $$(u - v)[4u^2 + 4uv + 4v^2 - 3] > 0$$ or $$4u^3 - 3u > 4v^3 - 3v$$ or (using triple angle formula $\cos 3x = 4\cos^3 x - 3\cos x$) $$\cos \left(3 \times \frac{50\pi}{180}\right) > \frac{45}{2} - \frac{27}{2}\sqrt 3$$ or $$13\sqrt 3 > \frac{45}{2}$$ which is true.
We are done.
A Taylor approach can be set up like this. Writing down a few Taylor terms of the expansion of $\cos$ around $\pi/4$ with Lagrange remainder, we get
$$\cos(5\pi/18)=\cos(\pi/4)+(-\sin(\pi/4)(\pi/36)+(-\cos(\pi/4))(\pi/36)^2/2 \\+\sin(\xi)(\pi/36)^3/6$$
for some $\xi \in (\pi/4,5\pi/18)$. Since $\sin$ is increasing on this interval, we get
$$\cos(5\pi/18)>(\sqrt{2}/2)(1-\pi/36-(\pi/36)^2/2+(\pi/36)^3/6).$$
Therefore we get
$$2\cos(5\pi/18)+\sqrt{3}>\sqrt{2}(1-\pi/36-(\pi/36)^2/2+(\pi/36)^3/6)+\sqrt{3}.$$
To polish off the problem, you can replace $\sqrt{2}$ by a lower bound of 1.41, $\pi$ by an upper bound of 3.15, and $\sqrt{3}$ by a lower bound of 1.73, which is enough accuracy to complete the proof. This is still a lot of arithmetic, although you can save some trouble by neglecting the last term entirely.
It is known that $\cos(50^o)=\frac{\phi}2$. So, we need to find the floor of $\phi+\sqrt{3}$. $\sqrt{3}$ is less than $2$ since it is less than $\sqrt{4}$. $\phi$ is less than $2$ since $(1+\sqrt{5})/2<(1+\sqrt{9})/2=2$. We could do a similar proof that $\sqrt{3},\phi>1$. So, the answer must be between $\lfloor4\rfloor$ and $\lfloor2\rfloor$. The answer is thus $3$.
