Question

Дано:
$\sin( \alpha ) \cos( \beta ) = \frac{1}{4}$
$\alpha + \beta = \frac{ - \pi}{6}$
Найди:

$4 \sin( \alpha - \beta ) =$
​

Answer 1

$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\sin\beta\cos\alpha=\frac14-\sin\beta\cos\alpha\\4\sin(\alpha-\beta)=1-4\sin\beta\cos\alpha\\\\\sin\beta\cos\alpha=\frac12\cdot\left(\sin(\beta-\alpha)+\sin(\beta+\alpha)\right)=\frac12\cdot\left(-\sin(\alpha-\beta)+\sin(\alpha+\beta)\right)=\\=\frac12\cdot\left(-\sin(\alpha-\beta)+\sin\left(-\frac\pi6\right)\right)=\frac12\cdot\left(-\sin(\alpha-\beta)-\frac12\right)$

$4\sin\beta\cos\alpha=4\cdot\frac12\cdot\left(-\sin(\alpha-\beta)-\frac12\right)=-2\sin(\alpha-\beta)-1\\\\\\4\sin(\alpha-\beta)=1+2\sin(\alpha-\beta)+1\\2\sin(\alpha-\beta)=2\\\boxed{4\sin(\alpha-\beta)=4}$