Question

27.8(1-4) срочно и спасибо кто откликнется

Answer 1

$1)\; \; sin^2\alpha -sin^2\beta=(sin\alpha -sin\beta )(sin\alpha +sin\beta )=\\\\=2\cdot sin\frac{\alpha -\beta }{2}\cdot cos\frac{\alpha +\beta }{2}\cdot 2\cdot sin\frac{\alpha +\beta}{2}\cdot cos\frac{\alpha -\beta }{2}=sin(\alpha -\beta )\cdot sin(\alpha +\beta )$

$2)\; \; cos^2\alpha -cos^2\beta=(cos\alpha -cos\beta )(cos\alpha +cos\beta )=\\\\=-2\cdot sin\frac{\alpha -\beta }{2}\cdot sin\frac{\alpha +\beta }{2}\cdot 2\cdot cos\frac{\alpha +\beta }{2}\cdot cos\frac{\alpha -\beta }{2}=-sin(\alpha -\beta )\cdot sin(\alpha +\beta )$

$3)\; \; \frac{3}{4}-sin^2x=(\frac{\sqrt3}{2}-sinx)(\frac{\sqrt3}{2}+sinx)=\\\\=(sin\frac{\pi}{3}-sinx)(sin\frac{\pi}{3}+sinx)=2\cdot sin\frac{\pi -3x}{6}\cdot cos\frac{\pi +3x}{6}\cdot 2\cdot sin\frac{\pi +3x}{6}\cdot cos\frac{\pi -3x}{6}=\\\\=sin\frac{\pi -3x}{3}\cdot sin\frac{\pi +3x}{3}=sin(\frac{\pi}{3}-x)\cdot sin(\frac{\pi}{3}+x)$

$4)\; \; cos^2x-\frac{1}{2}=(cosx-\frac{\sqrt2}{2})(cos+\frac{\sqrt2}{2})=\\\\=(cosx-cos\frac{\pi}{4})(cosx+cos\frac{\pi}{4})=-2\cdot sin\frac{4x-\pi}{8}\cdot sin\frac{4x+\pi }{8}\cdot 2\cdot cos\frac{4x+\pi}{8}\cdot cos\frac{4x-\pi}{8}=\\\\=-sin\frac{4x-\pi}{4}\cdot sin\frac{4x+\pi }{4}=-sin(x-\frac{\pi}{4})\cdot sin(x+\frac{\pi}{4})$