Question

Buna seara! Ma puteti ajuta cu aceasta problema?

Answer 1

Explicație pas cu pas:

$0 < x < \frac{\pi}{2}$

$\sin( \alpha ) + \sin( \beta ) = 2 \sin( \frac{ \alpha + \beta }{2} ) \cos( \frac{ \alpha - \beta }{2} )$

$\sin( \alpha ) - \sin( \beta ) = 2 \cos( \frac{ \alpha + \beta }{2} ) \sin( \frac{ \alpha - \beta }{2} )$

=>

$\frac{ \sin(3x) }{ \sin(x) } - \frac{ \cos(3x) }{ \cos(x) } = \\ = \frac{ \sin(3x) \cos(x) - \cos(3x) \sin(x) }{ \sin(x) \cos(x) } \\ = \frac{ \frac{\sin(4x) +\sin(2x) }{2} - \frac{\sin(4x) - \sin(2x) }{2}}{\sin(x) \cos(x)} \\ = \frac{\sin(4x) +\sin(2x) - \sin(4x) +\sin(2x)}{2\sin(x) \cos(x)} \\ = \frac{2\sin(2x)}{\sin(2x)} = 2$