Question

Sin a=3/5,a aparține (π/2;π) . Efectuați:a)sin2a b)tga. Cine mă ajută dau coroana. Va rog am nevoie urgentă

Answer 1

[tex]\text{Il aflam pe }\cos x\text{ dn teorema fundamentala a trigonometriei.}\\ \sin^2 x+\cos^2 x=1\\ \dfrac{9}{25}+\cos^2 x=1\\ \cos^2 x=1-\dfrac{9}{25}\\ \cos^2 x=\dfrac{16}{25}\\ \cos x=\pm \dfrac{4}{5}\\ x\in \left(\dfrac{\pi}{2},\pi\right)\Rightarrow \cos x=-\dfrac{4}{5}\\ \sin2x=2\sin x\cos x=2\cdot \dfrac{3}{5}\cdot \left(-\dfrac{4}{5}\right)=\boxed{-\dfrac{24}{25} }\\ tg\ x=\dfrac{\sin x}{\cos x}=\dfrac{\frac{3}{5}}{-\frac{4}{5}}=\boxed{-\dfrac{3}{4}} [/tex]