Question

Даю 40 баллов! Помогите пожалуйста ​

Answer 1

$1)\; \; \; \; 2sint\cdot sin2t+cos3t=2\cdot \frac{1}{2}\, (cost-cos3t)+cos3t=cost\\\\\\2)\; \; a)\; \; cos14^\circ \cdot cos16^\circ =\frac{1}{2}\, (cos30^\circ +cos2^\circ )=\frac{1}{2}\cdot (\frac{\sqrt3}{2}+cos2^\circ )=\frac{\sqrt3}{4}+\frac{cos2^\circ }{2}$

$b)\; \; sin^210^\circ +cos50^\circ \cdot cos70^\circ =sin^210^\circ +\frac{1}{2}\, (cos120^\circ +cos20^\circ )=\\\\=\frac{1-cos20^\circ }{2}+\frac{1}{2}\, (-cos\, 60^\circ +cos20^\circ)=\frac{1}{2}-\frac{1}{2}\, cos20^\circ -\frac{1}{2}\cdot \frac{1}{2}+\frac{1}{2}\, cos20^\circ =\frac{1}{4}$

$c)\; \; 3\, cos\frac{3\pi }{4}\cdot cos\frac{5\pi }{12}=\frac{3}{2}\, (cos\frac{7\pi}{6}+cos\frac{\pi}{3})=\frac{3}{2}\cdot (cos(\pi +\frac{\pi}{6})+\frac{1}{2})=\\\\=\frac{3}{2}\cdot (-cos\frac{\pi}{6}+\frac{1}{2})=\frac{3}{2}\cdot (\frac{1}{2}-\frac{\sqrt3}{2})$

$3)\; \; sina\cdot cos\beta =\frac{1}{4}\; \; ,\; \; \alpha +\beta =-\frac{\pi}{6}\\\\sin\alpha \cdot cos\beta =\frac{1}{2}\, (sin(\alpha +\beta )+sin(\alpha -\beta ))\\\\\frac{1}{4}=\frac{1}{2}\, (sin(-\frac{\pi}{6})+sin(\alpha -\beta ))\\\\\frac{2}{4}=-\frac{1}{2}+sin(\alpha -\beta )\\\\sin(\alpha -\beta )=1$