Question

даю 100 баллов срочно ​

Answer 1

1а)

$t=\frac{\pi}{6}=\frac{180^\circ}{6}=30^\circ\\\\sin(t)=sin(\frac{\pi}{6})=sin(30^\circ)=\frac{1}{2}\\\\cos(t)=cos(\frac{\pi}{6})=cos(30^\circ)=\frac{\sqrt{3}}{2}$

1б)

$t=\frac{\pi}{2}=\frac{180^\circ}{2}=90^\circ\\\\sin(t)=sin(\frac{\pi}{2})=sin(90^\circ)=1\\\\cos(t)=cos(\frac{\pi}{2})=cos(90^\circ)=0$

2а)

$tg(-\frac{5\pi}{6})=tg(\pi-\frac{5\pi}{6})=tg(\frac{\pi}{6})=tg(30^\circ)=\frac{sin(30^\circ)}{cos(30^\circ)}=\frac{1}{2}:\frac{\sqrt{3}}{2}=\frac{1}{\sqrt{3}}$

2б)

$(sin\frac{\pi}{3}-2cos\frac{\pi}{2}+tg\frac{11\pi}{6}) \div tg(-\frac{\pi}{4})=\\\\=(\frac{\sqrt{3}}{2}-2*0-tg\frac{\pi}{6}) \div (-tg\frac{\pi}{4})=\\\\=(\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{3}) \div (-1)=\\\\=-(\frac{3*\sqrt{3}}{3*2}-\frac{2*\sqrt{3}}{2*3})=\\\\=-\frac{\sqrt{3}}{6}$

3)

$0 < 1,3 < \frac{\pi}{2}\\sin(1,3)>0\\sin(-1,3)=-sin(1,3)\\sin(-1,3)<0\\\\\pi < 3,2 < \frac{3\pi}{2}\\sin(3,2)<0\\sin(-3,2)=-sin(-3,2)\\sin(-3,2)>0\\\\sin(-3,2)>sin(-1,3)$

4)

$tg(-\frac{9\pi}{7})=tg(\frac{5\pi}{7})\\\\\pi<\frac{5\pi}{7}<\frac{3\pi}{2}\\\\tg(-\frac{9\pi}{7})<0\\\\\\sin(\frac{7\pi}{11})>0\\\\\\\frac{\pi}{2}<2<\pi\\\\ctg(2)<0\\\\\\tg(-\frac{9\pi}{7})*sin(\frac{7\pi}{11})*ctg(2)=(-)\times(+)\times(-)=+.$