Question

Sin(x)+sin(1/π)=sin((1/π)+x)

Answer 1

Ответ:

Объяснение:

См фото

Answer 2

Ответ:

Объяснение:

sin(x)+sin(1/π)=sin((1/π)+x)

воспользуемся формулой суммы синусов в ьлевой части и синусом двойного угла в правой части.

слева получим .

2sin(х/2+1/2п)*cos(х/2-1/2π)=2sin(х/2+1/2п)*cos(х/2+1/2π)

2sin(х/2+1/2п)*cos(х/2-1/2π)-2sin(х/2+1/2п)*cos(х/2+1/2π)=0

2sin(х/2+1/2п)*(cos(х/2-1/2π)-cos(х/2+1/2π))=0

sin(х/2+1/2п)=0  

х/2+1/2п=пк  *2  

х+1/п=2пк

х= -1/п+2пк

или  

cos(х/2-1/2π)-cos(х/2+1/2π)=0

cos(х/2-1/2π)=cos(х/2+1/2π)

х/2-1/2π=х/2+1/2π         или          -х/2+1/2π=х/2+1/2π  

 0=п нет решений                        х=2пк

Ответ. х= -1/п+2пк   или х=2пк

Answer 3

Let's verify the given equation:

$\sin(x) + \sin(\frac{1}{\pi}) = \sin(\frac{1}{\pi} + x)$

First, we can use the sum-to-product formula for sines:

$\sin(a) + \sin(b) = 2\sin(\frac{a+b}{2})\cos(\frac{a-b}{2})$

Using this formula, we get:

$2\sin(\frac{x + \frac{1}{\pi}}{2})\cos(\frac{x - \frac{1}{\pi}}{2}) = \sin(\frac{1}{\pi} + x)$

Now, we know that $\sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b)$. Therefore, we can rewrite the equation as:

$2\sin(\frac{x}{2} + \frac{1}{2\pi})\cos(\frac{x}{2} - \frac{1}{2\pi}) = \sin(\frac{1}{\pi})\cos(x) + \cos(\frac{1}{\pi})\sin(x)$

We can use the fact that $\sin(\frac{1}{\pi}) = 0$ and $\cos(\frac{1}{\pi}) = 1$ to simplify the equation further:

$2\sin(\frac{x}{2} + \frac{1}{2\pi})\cos(\frac{x}{2} - \frac{1}{2\pi}) = \cos(x)\sin(x)$

This equation doesn't seem to hold generally for all values of $x$ because of the presence of the product $\cos(x)\sin(x)$ on the right-hand side. Therefore, the initial equation seems to be incorrect.