Trigonometry problems

Dec 2015
859
115
Earth
(1) If \(\displaystyle z_1 \neq z_2 \) and \(\displaystyle z_1 , z_2 \in (0,\pi )\).
Prove that \(\displaystyle \: \sin\left(\frac{z_1 +z_2 }{2}\right) > \frac{\sin(z_1 ) +\sin(z_2 )}{2}\).

(2) Find all pairs \(\displaystyle (x,y)\) such that \(\displaystyle x^2 +2x\cdot \cos(xy)+1=0\).

(3) Find the number of solutions of the equation \(\displaystyle \sin(x)=\frac{x}{n} \; \), where \(\displaystyle n\in \mathbb{N} \: , n\neq 1\).

(4) Prove that \(\displaystyle \: \tan(a_0 )\cdot \tan(a_1 ) \cdot ... \cdot \tan(a_n )\geq n^{1+n}\).
 
Last edited by a moderator:
Mar 2015
177
67
Universe 2.71828i3.14159
1. \(\displaystyle \sin\left( \frac{z_1 +z_2}{2}\right) > \frac{1}{2}(\sin(z_1) + \sin(z_2)) = \sin\left( \frac{z_1 +z_2}{2}\right) \times \cos\left( \frac{z_1 -z_2}{2}\right) \Rightarrow 1 > \cos\left( \frac{z_1 -z_2}{2}\right)\)
 
Last edited by a moderator:
  • Like
Reactions: 1 person
Mar 2015
177
67
Universe 2.71828i3.14159
2. It is easy. For any non-zero real number x, find the value of y. And it is

$$y= \pm \frac{1}{x} \times \arccos\left( \frac{-1-x^2}{2x}\right)$$

It is not a precise answer...
 
Last edited by a moderator:
  • Like
Reactions: 1 person
Jun 2019
445
237
USA
(3) Find the number of solutions of the equation \(\displaystyle \sin(x)=\frac{x}{n} \; \), where \(\displaystyle n\in \mathbb{N} \: , n\neq 1\).
$\displaystyle N = 4 \left( \left\lceil \frac{n-1}{2\pi} \right\rceil - \varepsilon_n \right) - 1$
where $\varepsilon_n \in \mathbb{Z}$ is an approximation error that is usually zero for small $n$ but is nonzero increasingly more frequently as $n$ increases. :spin:
 
Last edited by a moderator:
  • Like
Reactions: 1 person
Dec 2015
859
115
Earth
By inspecting problem with a similar example, I got (4) :
(4) \(\displaystyle \tan(x_i ) < x_i \: \Rightarrow \prod_{i=1}^{n} \tan(x_i ) < \prod_{i=1}^{n} x_i \) .
Let \(\displaystyle x\) denote the largest number in set S:{\(\displaystyle x_1 , x_2 ,x_n \)}.
\(\displaystyle \prod_{i=1}^{n} \tan(x_i ) < \prod_{i=1}^{n} x_i < \underbrace{x\cdot x \cdot .... \cdot x }_{n} =\lfloor x^{n} \rfloor \leq n^n \).
 
Last edited by a moderator:
Dec 2015
859
115
Earth
$\displaystyle N = 4 \left( \left\lceil \frac{n-1}{2\pi} \right\rceil - \varepsilon_n \right) - 1$
where $\varepsilon_n \in \mathbb{Z}$ is an approximation error that is usually zero for small $n$ but is nonzero increasingly more frequently as $n$ increases. :spin:
Can we have a full solution ?