Question

Срочноооо все баллы!!!

Answer 1

$1)\; \; sina=-0,8\\\\\frac{3\pi }{2} <a<2\pi \; \; \to \; \; cosa<0\; ,\; \; tga>0\; ,\; \; ctga>0\\\\cosa=-\sqrt{1-sin^2a}=-\sqrt{1-0,64}=-0,6\\\\1+tg^2a=\frac{1}{cos^2a}\; \; \to \; \; tg^2a=\frac{1}{cos^2a}-1=\frac{1}{0,36}-1=\frac{0,64}{0,36}\; ,\\\\tga=+\frac{0,8}{0,6}=+\frac{4}{3}\\\\ctga=\frac{1}{tga}=+\frac{3}{4}\\\\2)\; \; cos(a+\beta )-cos(a-\beta )=-2\, sin\frac{a+\beta +a-\beta }{2}\cdot sin\frac{a+\beta -a+\beta }{2}=\\\\=-2\, sina\cdot sin\beta$

$3)\; \; sin(\frac{\pi}{3}+a)-sin(\frac{2\pi }{3}-a)=2\, sin\frac{\frac{\pi}{3}+a-\frac{2\pi }{3}+a }{2}\cdot cos\frac{\frac{\pi}{3}+a+\frac{2\pi }{3}-a}{2}=\\\\=2\, sin(a-\frac{\pi}{3})\cdot \underbrace {cos\pi }_{-1}=-2\, sin(a-\frac{\pi }{3})\\\\4)\; \; \frac{1-ctg^2a}{sin^4a-cos^4a}=\frac{1-\frac{cos^2a}{sin^2a}}{(sin^2a-cos^2a)(sin^2a+cos^2a)}=\frac{sin^2a-cos^2a}{sin^2a\cdot (sin^2a-cos^2a)\cdot 1}=\\\\=\frac{1}{sin^2a}\; \; ,\; \; \; \; \; (\frac{1}{sin^2a}=1+ctg^2a)$