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Fie x€ (π/2,π) astfel incat sin x=5/13 aflati cosx; tgx; ctgx; sin 2x

Răspuns :

[tex]\displaystyle sin~x= \frac{5}{13} ~~~~~~~~~~cos~x,~tg~x,~ctg~x,~sin~2x=? \\ \\ sin^2x+cos^2x=1 \Rightarrow cos^2x=1-sin^2x \Rightarrow cos~x=\pm \sqrt{1-sin^2x} \\ \\ x \in \left( \frac{\pi}{2} ,\pi \right) \Rightarrow cos x=- \sqrt{1-sin^2x} [/tex]
[tex]\displaystyle cos x=- \sqrt{1-sin^2x} \Rightarrow cos~x=- \sqrt{1-\left( \frac{5}{13} \right)^2} \Rightarrow \\ \\ \Rightarrow cos~x= -\sqrt{1- \frac{25}{169} } \Rightarrow cos~x=- \sqrt{ \frac{169-25}{169} } \Rightarrow \\ \\ \Rightarrow cos~x=- \sqrt{\frac{144}{169} } \Rightarrow cos~x= -\frac{ \sqrt{144} }{ \sqrt{169} } \Rightarrow cos~x=- \frac{12}{13} [/tex]
[tex]\displaystyle tg~x=\frac{sin~x}{cos~x} \Rightarrow tg~x=\frac{ \frac{5}{13} }{ -\frac{12}{13} } \Rightarrow tg~x=\frac{5}{13}:\left(- \frac{12}{13} \right) \Rightarrow \\ \\ \Rightarrow tg~x=\frac{5}{13} \cdot \left(- \frac{13}{12} \right)\Rightarrow tg~x=- \frac{5}{12} \\ \\ ctg~x= \frac{cos~x}{sin~x}\Rightarrow ctg~x= \frac{- \frac{12}{13} }{ \frac{5}{13}}\Rightarrow ctg~x=-\frac{12}{13}:\frac{5}{13} \Rightarrow\\ \\ \Rightarrow ctg~x=-\frac{12}{13} \cdot \frac{13}{5} \Rightarrow ctg~x=- \frac{12}{5}[/tex]
[tex]\displaystyle sin~2x=sin(x+x)=sin~x \cdot cos~x+sin~x \cdot~cosx= \\ \\ =2sin~x \cdot cos~x=2 \cdot \frac{5}{13} \cdot \left(-\frac{12}{13}\right) = \frac{10}{13} \cdot \left(-\frac{12}{13}\right) =- \frac{120}{169} [/tex]