Question

Тригонометрия:
Известно, что ctgx=3/4, x принадлежит [П;3П/2] вычисл: cos/sin/tg/ctg x/2

Answer 1

$ctg~x=\dfrac 34;\ \ \ \ x\in\bigg(\pi;\dfrac{3\pi}2\bigg)$       - III четверть

III  четверть :  cos x < 0;  sin x < 0;  tg x > 0

$ctg~x=\dfrac34\ \ \ \Rightarrow\ \ \ \ tg~x=\dfrac 43\\\\tg^2x+1=\dfrac1{\cos^2 x}\ \ \ \Rightarrow\ \ \ \cos^2x=\dfrac1{tg^2x+1}\\\\\cos x=-\sqrt{\dfrac 1{tg^2x+1}}=-\sqrt{\dfrac 1{\big(\frac43\big)^2+1}}=-\sqrt{\dfrac 9{25}}\\\\\boldsymbol{\cos x=-0,6}$

$\pi<x\leq\dfrac{3\pi}2\ \ \ \ \ \bigg|:2\\\\\dfrac{\pi}2<\dfrac x2\leq\dfrac{3\pi}4$

II  четверть :  

$\cos \dfrac x2 < 0;\ \ \sin \dfrac x2 > 0; \ \ \ tg ~\dfrac x2 < 0;\ \ \ ctg~\dfrac x2 < 0$

$\boldsymbol{\sin\dfrac x2=}\sqrt{\dfrac{1-\cos x}2}=\sqrt{\dfrac{1-(-0,6)}2}\boldsymbol{=\sqrt{\big{0,8}}}\\\\\\\boldsymbol{\cos\dfrac x2=}-\sqrt{\dfrac{1+\cos x}2}=-\sqrt{\dfrac{1-0,6}2}\boldsymbol{=-\sqrt{\big{0,2}}}\\\\\\\boldsymbol{tg ~\dfrac x2=}\dfrac{\sin \frac x2}{\cos \frac x2}=\dfrac{\sqrt{0,8}}{-\sqrt{0,2}}=-\sqrt{\dfrac 82}\boldsymbol{=-2}\\\\\\\boldsymbol{ctg~\dfrac x2=}\dfrac 1{tg~\frac x2}=\dfrac 1{-2}\boldsymbol{=-0,5}$