Question

27 БАЛЛОВ СРОЧНО!!!!

Answer 1

$7)\; \; \frac{4}{\sqrt3}\cdot sin(-420^\circ )=-\frac{4}{\sqrt3}\cdot sin(360^\circ +60^\circ )=-\frac{4}{\sqrt3}\cdot sin60^\circ =\\\\=-\frac{4}{\sqrt3}\cdot \frac{\sqrt3}{2}=-2\\\\\\8)\; \; (cos\frac{3\pi}{8}-sin\frac{3\pi}{8})(cos\frac{3\pi}{8}+sin\frac{3\pi}{8})=cos^2\frac{3\pi}{8}-sin^2\frac{3\pi}{8}=\\\\=cos\frac{3\pi}{4}=cos(\pi -\frac{\pi}{4})=-cos\frac{\pi}{4}=-\frac{\sqrt2}{2}$

$9)\; \; \frac{sin40^\circ}{sin20^\circ }=\frac{2\cdot sin20^\circ \cdot cos20^\circ }{sin20^\circ }=2\cdot cos20^\circ$

$10)\; \; -4(sin^270^\circ +sin^2160^\circ )=-4\, \Big(\frac{1-cos140^\circ }{2}+\frac{1-cos320^\circ }{2}\Big)=\\\\=-4\cdot \frac{1-cos(180^\circ -40^\circ )+1-cos(360^\circ -40^\circ )}{2}=\\\\=-2\cdot \Big(2+cos40^\circ -cos40^\circ \Big)=-2\cdot 2=-4$