Question

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

Answer 1

$sin(frac{pi}{22})-sin(frac{3pi}{22})+sin(frac{5pi}{22})-sin(frac{7pi}{22})+sin(frac{9pi}{22})=sin(frac{pi}{22})+(sin(frac{5pi}{22})-sin(frac{3pi}{22}))+(sin(frac{9pi}{22})-sin(frac{7pi}{22}))=\\=sin(frac{pi}{22})+2sin(frac{pi}{22})cos(frac{2pi}{11})+2sin(frac{pi}{22})cos(frac{4pi}{11})=sin(frac{pi}{22})times(1+2cos(frac{2pi}{11})+2cos(frac{4pi}{11}))=\\=sin(frac{pi}{22})times(1+2(cos(frac{2pi}{11})+cos(frac{4pi}{11})))=sin(frac{pi}{22})times(1+4cos(frac{3pi}{11})cos(frac{pi}{11}))=\\=frac{2cos(frac{pi}{22})timessin(frac{pi}{22})times(1+4cos(frac{3pi}{11})cos(frac{pi}{11}))}{2cos(frac{pi}{22})}=frac{sin(frac{pi}{11})times(1+4cos(frac{3pi}{11})cos(frac{pi}{11}))}{2cos(frac{pi}{22})}=\\=frac{sin(frac{pi}{11})+sin(frac{pi}{11})times4cos(frac{3pi}{11})cos(frac{pi}{11})}{2cos(frac{pi}{22})}=frac{sin(frac{pi}{11})+2cos(frac{3pi}{11})sin(frac{2pi}{11})}{2cos(frac{pi}{22})}=\\=frac{sin(frac{pi}{11})-sin(frac{pi}{11})+sin(frac{5pi}{11})}{2cos(frac{pi}{22})}=frac{sin(frac{5pi}{11})}{2cos(frac{pi}{22})}=frac{cos(frac{pi}{2}-frac{5pi}{11})}{2cos(frac{pi}{22})}=frac{cos(frac{pi}{22})}{2cos(frac{pi}{22})}=frac{1}{2}=0.5$

Основные формулы:

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

ОТВЕТ: 0,5