Without Calculator find $$\left\lfloor 2 \cos \left(50^{\circ}\right)+\sqrt{3}\right\rfloor$$
Where $\left \lfloor x \right \rfloor $ represents floor function.
My Try:
Let $x=2\cos(50^{\circ})+\sqrt{3}$. We have
$$\begin{aligned}
& \cos \left(50^{\circ}\right)<\cos \left(45^{\circ}\right) \\
\Rightarrow \quad & 2 \cos \left(50^{\circ}\right)<\sqrt{2} \\
\Rightarrow \quad & x<\sqrt{3}+\sqrt{2}<3.14
\end{aligned}$$
Now I am struggling hard to prove that $x>3$
Best Answer
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: $45^2=26^2 \times 3-3$.