Question

1.в какой координатной четверти лежит радиус-вектор, соответствующий угол: 1)740° 2)-520° 3)7п/3 4)-11п/4 2. 1)sin300°×cos300°×tg290° 2)sin3п/4×cos3п/4×ctg4п/3 : )

Answer 1

$1)\; \; 740^\circ =2\cdot 360^\circ +20^\circ \quad -\; \; 1\; chetvert\\\\2)\; \; -520^\circ =-360^\circ -160^\circ \quad -\; \; 3\; chetvert\\\\3)\; \; \frac{7\pi }{3}= \frac{3\cdot 2\pi +\pi }{3}= 2\pi +\frac{\pi }{3} \quad -\; \; 1\; chetvert\\\\4)\; \; - \frac{11\pi }{4} =\frac{-8\pi -3\pi }{4} =-2\pi -\frac{3\pi }{4}\quad -\; \; 3\; chetvert$

$5)\; \; sin300^\circ \cdot cos300^\circ \cdot tg290^\circ =\\\\=sin(360^\circ -60^\circ )\cdot cos(360^\circ -60^\circ )\cdot tg(270^\circ +20^\circ )=\\\\=-sin60^\circ \cdot cos60^\circ \cdot (-ctg20^\circ )=\frac{1}{2}sin120^\circ \cdot ctg20^\circ =\\\\=\frac{1}{2}sin(180^\circ -60^\circ )\cdot ctg20^\circ =\frac{1}{2}sin60^\circ \cdot ctg20^\circ =\frac{1}{2}\cdot \frac{\sqrt3}{2}\cdot ctg20^\circ =\\\\=\frac{\sqrt3}{4}\cdor ctg20^\circ$

$6)\; \; sin \frac{3\pi }{4} \cdot cos \frac{3\pi }{4}\cdot ctg \frac{4\pi }{3} = \frac{1}{2}sin \frac{3\pi }{2}\cdot ctg(\pi +\frac{\pi }{3})=\\\\= \frac{1}{2}\cdot (-1)\cdot ctg\frac{\pi }{3}= -\frac{1}{2}\cdot \frac{\sqrt3}{3} = -\frac{\sqrt3}{6}$

Answer 2

1
740=360*2+20 1 ч
-520=-360-160  3ч
7π/3=2π+π/3  1ч
-11π/4=-2π-3π/4  3ч
2
sin(360-60)*cos(360-60)*tg(270+20)--sin60*cos60*(-сtg20)=
=-√3/2*1/2*(-сtg20)=√3/4*сtg20

sin(π-π/4)*cos(π-π/4)*ctg(π+π/3)=sinπ/4*(-cosπ/4)*ctgπ/3=
=1/√2*(-1/√2)*√3/3=-√3/6