Help me. Срочно надо, пожалуйста помогите!:(

Question 1

Help me. Срочно надо, пожалуйста помогите!:(

Question 2

$ctg(t-\pi )=-ctg(\pi-t)=+ctgt=-\frac{3}{4}\\\\cos(\frac{3\pi}{2}-t)=-sint,\; \; cos(\pi +t)=-cost\\\\Formyla:\; \; 1+ctg^2t=\frac{1}{sin^2t}\; ,\; \to \; sin^2t=\frac{1}{1+ctg^2t}=\frac{1}{1+\frac{9}{16}}=\frac{16}{25}$
0,\; cost<0 \\\\sint=+\frac{4}{5},\; \; cost=-\sqrt{1-sin^2t}=-\sqrt{1-\frac{16}{25}}=-\sqrt{\frac{9}{25}}=-\frac{3}{5}\\\\Otvet:\; cos(\frac{3\pi}{2}-t)=-sint=-\frac{4}{5},\; \; cos(\pi +t)=-cost=\frac{3}{5}\\\\2)\; d<c<b<a" alt="Tak\; \; kak\; \; \frac{\pi}{2}<t<\pi ,to\; sint>0,\; cost<0 \\\\sint=+\frac{4}{5},\; \; cost=-\sqrt{1-sin^2t}=-\sqrt{1-\frac{16}{25}}=-\sqrt{\frac{9}{25}}=-\frac{3}{5}\\\\Otvet:\; cos(\frac{3\pi}{2}-t)=-sint=-\frac{4}{5},\; \; cos(\pi +t)=-cost=\frac{3}{5}\\\\2)\; d<c<b<a" align="absmiddle" class="latex-formula">

Question 3

1+ctg^2(π-t) =1/sin^2(π-t) 1+9/16=1/ sin^2(π-t) sin^2(π-t)=16/25 sin(π-t)=0,8 так как π-t при π/2 < t <π наход-ся в первой четверти sin t=0,8 cos t=-0,6 cos(3π/2-t)=-sin t=-0,8 c0s(π+t)=-cos t=0,6 1) sin4 2) sin 6 3) cos7 4) cos6

NNNLLL54_zn · Answer 1 · 2018-03-15T00:16:59+0000

$ctg(t-\pi )=-ctg(\pi-t)=+ctgt=-\frac{3}{4}\\\\cos(\frac{3\pi}{2}-t)=-sint,\; \; cos(\pi +t)=-cost\\\\Formyla:\; \; 1+ctg^2t=\frac{1}{sin^2t}\; ,\; \to \; sin^2t=\frac{1}{1+ctg^2t}=\frac{1}{1+\frac{9}{16}}=\frac{16}{25}$
0,\; cost<0 \\\\sint=+\frac{4}{5},\; \; cost=-\sqrt{1-sin^2t}=-\sqrt{1-\frac{16}{25}}=-\sqrt{\frac{9}{25}}=-\frac{3}{5}\\\\Otvet:\; cos(\frac{3\pi}{2}-t)=-sint=-\frac{4}{5},\; \; cos(\pi +t)=-cost=\frac{3}{5}\\\\2)\; d<c<b<a" alt="Tak\; \; kak\; \; \frac{\pi}{2}<t<\pi ,to\; sint>0,\; cost<0 \\\\sint=+\frac{4}{5},\; \; cost=-\sqrt{1-sin^2t}=-\sqrt{1-\frac{16}{25}}=-\sqrt{\frac{9}{25}}=-\frac{3}{5}\\\\Otvet:\; cos(\frac{3\pi}{2}-t)=-sint=-\frac{4}{5},\; \; cos(\pi +t)=-cost=\frac{3}{5}\\\\2)\; d<c<b<a" align="absmiddle" class="latex-formula">