Question

Тригонометрия.
Найти значение выражения:
$sin(\frac{\pi }{22})-sin(\frac{3\pi }{22})+sin(\frac{5\pi }{22})-sin(\frac{7\pi }{22})+sin(\frac{9\pi }{22})$

Answer 1

$\sin(\frac{\pi}{22})-\sin(\frac{3\pi}{22})+\sin(\frac{5\pi}{22})-\sin(\frac{7\pi}{22})+\sin(\frac{9\pi}{22})=\sin(\frac{\pi}{22})+(\sin(\frac{5\pi}{22})-\sin(\frac{3\pi}{22}))+(\sin(\frac{9\pi}{22})-\sin(\frac{7\pi}{22}))=\\\\=\sin(\frac{\pi}{22})+2\sin(\frac{\pi}{22})\cos(\frac{2\pi}{11})+2\sin(\frac{\pi}{22})\cos(\frac{4\pi}{11})=\sin(\frac{\pi}{22})\times(1+2\cos(\frac{2\pi}{11})+2\cos(\frac{4\pi}{11}))=\\\\=\sin(\frac{\pi}{22})\times(1+2(\cos(\frac{2\pi}{11})+\cos(\frac{4\pi}{11})))=\sin(\frac{\pi}{22})\times(1+4\cos(\frac{3\pi}{11})\cos(\frac{\pi}{11}))=\\\\=\frac{2\cos(\frac{\pi}{22})\times\sin(\frac{\pi}{22})\times(1+4\cos(\frac{3\pi}{11})\cos(\frac{\pi}{11}))}{2\cos(\frac{\pi}{22})}=\frac{\sin(\frac{\pi}{11})\times(1+4\cos(\frac{3\pi}{11})\cos(\frac{\pi}{11}))}{2\cos(\frac{\pi}{22})}=\\\\=\frac{\sin(\frac{\pi}{11})+\sin(\frac{\pi}{11})\times4\cos(\frac{3\pi}{11})\cos(\frac{\pi}{11})}{2\cos(\frac{\pi}{22})}=\frac{\sin(\frac{\pi}{11})+2\cos(\frac{3\pi}{11})\sin(\frac{2\pi}{11})}{2\cos(\frac{\pi}{22})}=\\\\=\frac{\sin(\frac{\pi}{11})-\sin(\frac{\pi}{11})+\sin(\frac{5\pi}{11})}{2\cos(\frac{\pi}{22})}=\frac{\sin(\frac{5\pi}{11})}{2\cos(\frac{\pi}{22})}=\frac{\cos(\frac{\pi}{2}-\frac{5\pi}{11})}{2\cos(\frac{\pi}{22})}=\frac{\cos(\frac{\pi}{22})}{2\cos(\frac{\pi}{22})}=\frac{1}{2}=0.5$

Основные формулы:

$1)\:\:\sin(\alpha )-\sin(\beta)=2\sin(\frac{\alpha-\beta}{2})\cos(\frac{\alpha+\beta}{2})\\2)\:\:\cos(\alpha)+\cos(\beta)=2\cos(\frac{\alpha+\beta}{2})\cos(\frac{\alpha-\beta}{2})\\3)\:\:2\sin(\alpha)\cos(\alpha)=\sin(2\alpha)\\4)\:\:2\sin(\alpha)\cos(\beta)=\sin(\alpha-\beta)+\sin(\alpha+\beta)\\5)\sin(\alpha)=\cos(\frac{\pi}{2}-\alpha)$

ОТВЕТ: 0,5