Question

Se considera x apartine (3pi/2;2pi),astfel incat sinx+cosx= (1-radical din 3)/2. Sa se determine sinx, cosx, tgx si ctg x.
Va rog!!! E urgent! Dau coroana

Answer 1

[tex]\it sinx + cosx = \dfrac{1-\sqrt3}{2}\ \ \ \ \ (1) \\ \\ \\ (1) \Rightarrow (sinx+cosx)^2 = \left(\dfrac{1-\sqrt3}{2}\right)^2 \Rightarrow sin^2x+cos^2x +2sinxcosx = \\ \\ \\ \left(\dfrac{1-\sqrt3}{2}\right)^2 \Rightarrow 1 +2sinxcosx = \left(\dfrac{1-\sqrt3}{2}\right)^2 \Rightarrow 2sinxcosx = \\ \\ \\ = \left(\dfrac{1-\sqrt3}{2}\right)^2 -1 \Rightarrow 2sinxcosx = \left(\dfrac{1-\sqrt3}{2} -1\right)\left(\dfrac{1-\sqrt3}{2} +1\right) \Rightarrow [/tex]

[tex]\it \Rightarrow 2sinxcosx= \dfrac{1-\sqrt3-2}{2}\cdot\dfrac{1-\sqrt3+2}{2} = - \dfrac{(\sqrt3+1)(3-\sqrt3)}{4} = \\ \\ \\ =-\dfrac{3\sqrt3-3+3-\sqrt3}{4} =- \dfrac{\sqrt3}{2} \Rightarrow \\ \\ \\ \Rightarrow sinxcosx= -\dfrac{\sqrt3}{4} \ \ \ \ \ (2)[/tex]

$\it x\in \left(\dfrac{3\pi}{2},\ 2\pi\right) \Rightarrow sinx\ \textless \ 0,\ cosx\ \textgreater \ 0 \ \ \ \ \ (3)$

(1), (2) ⇒ sinx,  cosx  sunt rădădcinile ecuației:

$\it t^2+\dfrac{\sqrt3-1}{2}t - \dfrac{\sqrt3}{4} =0 \Rightarrow 4t^2+2(\sqrt3-1)t-\sqrt3 =0$

După rezolvarea ultimei ecuații, rezultă:

[tex]\it t_1 = -\dfrac{\sqrt3}{2}\ \textless \ 0 \stackrel{(3)}{\Longrightarrow} sinx = -\dfrac{\sqrt3}{2} \\ \\ \\ t_2 = \dfrac{1}{2}\ \textgreater \ 0 \stackrel{(3)}{\Longrightarrow} cosx = \dfrac{1}{2} [/tex]


[tex]\it tgx = \dfrac{sinx}{cosx} = \dfrac{-\dfrac{\sqrt3}{2}}{\dfrac{1}{2}} = -\dfrac{\sqrt3}{2} \cdot\dfrac{2}{1} = -\sqrt3 \\ \\ \\ ctgx = \dfrac{1}{tgx} = -\dfrac{1}{\sqrt3} =-\dfrac{\sqrt3}{3}[/tex]