# Trigonometry problems

#### idontknow

(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 $$\displaystyle \: \tan(a_0 )\cdot \tan(a_1 ) \cdot ... \cdot \tan(a_n )\geq n^{1+n}$$.

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#### tahirimanov19

1. $$\displaystyle \sin\left( \frac{z_1 +z_2}{2}\right) > \frac{1}{2}(\sin(z_1) + \sin(z_2)) = \sin\left( \frac{z_1 +z_2}{2}\right) \times \cos\left( \frac{z_1 -z_2}{2}\right) \Rightarrow 1 > \cos\left( \frac{z_1 -z_2}{2}\right)$$

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#### tahirimanov19

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...

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#### tahirimanov19

And 4. is wrong...

#### DarnItJimImAnEngineer

(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:

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#### skipjack

Forum Staff
2. It is easy.
. . .
It is not a precise answer...
For $y$ to be real, $x$ must be 1 or -1. Now it's easy.

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#### idontknow

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:{$$\displaystyle x_1 , x_2 ,x_n$$}.
$$\displaystyle \prod_{i=1}^{n} \tan(x_i ) < \prod_{i=1}^{n} x_i < \underbrace{x\cdot x \cdot .... \cdot x }_{n} =\lfloor x^{n} \rfloor \leq n^n$$.

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#### idontknow

$\displaystyle N = 4 \left( \left\lceil \frac{n-1}{2\pi} \right\rceil - \varepsilon_n \right) - 1$
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 ?