Question

$1)sin14x*cos10x=\frac{1}{2} sin24x\\\\2)sin9x*sin3x=\frac{1}{2}cos6x$

Answer 1

Ответ:

1.

$\sin(14x) \times \cos(10x) = \frac{1}{2} \sin(24x) \\ \sin(14x) \cos(10 x) = \frac{1}{2} \sin(14x + 10x) \\ \sin(14x) \cos(10x) = \frac{1}{2} ( \sin(14x) \cos(10x) + \sin(10x) \cos(14x) ) \\ \sin(14x) \cos(10x) - \frac{1}{2} \sin(14x) \cos(10x) - \frac{1}{2} \sin(10x) \cos(14x) = 0\\ \frac{1}{2} \sin(14x) \cos(10x) - \frac{1}{2} \sin(10x) \cos(14x) = 0 \\ \frac{1}{2} ( \sin(14x) \cos(10x) - \sin(10x) \cos(14x) ) = 0 \\ \frac{1}{2} \sin(14x - 10x) = 0 \\ \sin(4x) = 0 \\ 4x = \pi \: n \\ x = \frac{\pi \: n}{4}$

n принадлежит Z.

2.

$\sin(9x) \sin(3x) = \frac{1}{2} \cos(6x) \\ \sin(9x) \sin(3x) - \frac{1}{2} \cos(9x - 3x) = 0 \\ \sin(9x) \sin(3x) - \frac{1}{2} ( \cos(9x) \cos(3x) + \sin(9x) \sin(3x) ) = 0 \\ \frac{1}{2} \sin(9x) \sin(3x) - \frac{1}{2} \cos(9x) \cos(3x) = 0 \\ - \frac{1}{2} ( \cos(9x) \cos(3x) - \sin(9x) \sin(3x) ) = 0 \\ - \frac{1}{2} \cos(9x + 3x) = 0 \\ \cos(12x) = 0 \\ 12x = \frac{\pi}{2} + 2 \pi \: n \\ x = \frac{\pi}{24} + \frac{\pi \: n}{6}$

n принадлежит Z.