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

Question 1

Question 2

какой вариант?

Question 3

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