Question

Привет. Помогите решить С-6 (домашняя самостоятельная работа).

Answer 1

1)  а)  $sin18sin54=sin18cos36* frac{2cos18}{2cos18} =frac{sin36cos36}{2cos18}* frac{2}{2} =$ 
$=frac{sin72}{4cos18}=frac{cos18}{4cos18}=frac{1}{4}$ 
б)  $cosfrac{2pi}{7} +cosfrac{4pi}{7}+cosfrac{6pi}{7} =cosfrac{2pi}{7}+cosfrac{6pi}{7}+cosfrac{4pi}{7}=$
$=-2cosfrac{4pi}{7}sinfrac{2pi}{7}* frac{cosfrac{2pi}{7}}{cosfrac{2pi}{7}} +cosfrac{4pi}{7}=$
$=- frac{cosfrac{4pi}{7}sinfrac{4pi}{7}}{cosfrac{2pi}{7}} * frac{2}{2} +cosfrac{4pi}{7}=$
$=-frac{sinfrac{8pi}{7}}{2cosfrac{2pi}{7}}+cosfrac{4pi}{7}=-frac{sin( pi +frac{pi}{7})}{2cosfrac{2pi}{7}}+cosfrac{4pi}{7}=$
$=-frac{-sinfrac{pi}{7}}{2cosfrac{2pi}{7}}+cosfrac{4pi}{7}=frac{sinfrac{pi}{7}}{2cosfrac{2pi}{7}}+cosfrac{4pi}{7}=$
2 )а)  $sin^{2}(frac{9pi }{8}+alpha)-sin^{2}(frac{15pi }{8}+alpha)=$
$(sin(frac{9pi }{8}+alpha)-sin(frac{15pi }{8}+alpha))(sin(frac{9pi }{8}+alpha)+sin(frac{15pi }{8}+alpha))=$
$=(-2sin(3pi +alpha)sin(frac{-3pi }{4}))*2sin(3pi +alpha)cos(frac{-3pi }{4})=$$=4sinalpha(-frac{sqrt{2}}{2})(-sinalpha)(-frac{sqrt{2}}{2})=-sin2 alpha$
б)  $sin^{2}2 alpha -sin^{2} beta +cos(2 alpha + beta )cos(2 alpha - beta )=$$=(sin2 alpha -sinbeta )(sin2 alpha +sinbeta )+cos(2 alpha + beta )cos(2 alpha - beta )=$
$=2sinfrac{2alpha-beta}{2}cosfrac{2alpha+beta}{2}*2sinfrac{2alpha+beta}{2}cosfrac{2alpha-beta}{2}+cos(2 alpha + beta )cos(2 alpha - beta )=$
$=sin(2alpha-beta)sin(2alpha+beta)+cos(2 alpha + beta )cos(2 alpha - beta )=$
$=cos2 beta$