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Fie x apartine (pi/2, pi) si cos2x=-1/2.Se cer cos x si sin x.

Răspuns :

[tex]x \: \in \: ( \frac{\pi}{2} ,\pi) = cadranul \: II = > sin \: x > 0,cos \: x < 0[/tex]

[tex]cos2x = - \frac{1}{2} [/tex]

[tex]cos2x = {cos}^{2} x - {sin}^{2} x[/tex]

[tex]\left\{\begin{matrix}

{cos}^{2}x - {sin}^{2}x = - \frac{1}{2} \\ {sin}^{2} x + {cos}^{2} x = 1

\end{matrix}\right.[/tex]

[tex] {cos}^{2} x - {sin}^{2} x + {sin}^{2} x + {cos}^{2} x = - \frac{1}{2} + 1[/tex]

[tex]2 {cos}^{2} x = \frac{1}{2} [/tex]

[tex] {cos}^{2} x = \frac{1}{4} [/tex]

[tex]cos \: x = \pm \: \sqrt{ \frac{1}{4} } = \pm \: \frac{1}{2} [/tex]

[tex]x \: \in \: ( \frac{\pi}{2} ,\pi) = > cos \: x = - \frac{1}{2} [/tex]

[tex] {sin}^{2} x + {cos}^{2} x = 1[/tex]

[tex] {sin}^{2} x + \frac{1}{4} = 1[/tex]

[tex] {sin}^{2} x = 1 - \frac{1}{4} [/tex]

[tex] {sin}^{2} x = \frac{3}{4} [/tex]

[tex]sin \: x = \pm \: \sqrt{ \frac{3}{4} } = \pm \: \frac{ \sqrt{3} }{ \sqrt{4} } = \pm \: \frac{ \sqrt{3} }{2} [/tex]

[tex]x \: \in \: ( \frac{\pi}{2} ,\pi) = > sin \: x = \frac{ \sqrt{3} }{2} [/tex]