[tex]\\ $Il scoatem pe $ \sin\Big(\dfrac{x}{2}\Big).\\ \\ \cos2x = \cos^2 x-\sin^2 x\Rightarrow \cos2x = (1-\sin^2 x)-\sin^2 x \Rightarrow \\ \\ \Rightarrow \cos2x = 1-\sin^2 x-\sin^2 x \Rightarrow \cos2x = 1-2\sin^2 x \Rightarrow \\ \\ \Rightarrow 2\sin^2 x = 1 - \cos2x \Rightarrow \sin^2 x = \dfrac{1-\cos2x}{2} \Rightarrow \\ \\ \Rightarrow \sin^2 \Big(\dfrac{x}{2}\Big) = \dfrac{1-\cos \Big(2\cdot \dfrac{x}{2}\Big)}{2} \Rightarrow \sin^2 \Big(\dfrac{x}{2}\Big) = \dfrac{1-\cos x}{2}[/tex]
[tex]\\ $Avem nevoie de $ \cos x.\\ \\ \sin^2x+\cos^2x = 1 \Rightarrow \cos^2x = 1-\sin^2x \Rightarrow \cos^2 x = 1-\Big(\dfrac{3}{5}\Big)^2 \Rightarrow \\ \\ \Rightarrow \cos^2x = 1 - \dfrac{9}{25} \Rightarrow \cos^2 x = \dfrac{25-9}{25} \Rightarrow \cos^2 x = \dfrac{16}{25} \Rightarrow \\ \\ \Rightarrow\cos x = \pm \sqrt{\dfrac{16}{25}} \Rightarrow \cos x = \dfrac{4}{5},$ dar, $ x\in \Big(\dfrac{\pi}{2},\pi\Big) \Rightarrow \cos x = -\dfrac{4}{5}[/tex]
[tex]\sin^2 \Big(\dfrac{x}{2}\Big) = \dfrac{1-\Big(-\dfrac{4}{5}\Big)}{2} \Rightarrow \sin^2 \Big(\dfrac{x}{2}\Big) =\dfrac{1+\dfrac{4}{5}}{2} \Rightarrow \sin^2 \Big(\dfrac{x}{2}\Big) =\dfrac{\dfrac{5+4}{5}}{2} \Rightarrow \\ \\ \Rightarrow \sin^2 \Big(\dfrac{x}{2}\Big) =\dfrac{5+4}{2\cdot 5} \Rightarrow \sin^2 \Big(\dfrac{x}{2}\Big) =\dfrac{9}{10} \Rightarrow \sin \Big(\dfrac{x}{2}\Big) =\pm\sqrt{\dfrac{9}{10}} ,$ dar, $ \\ \\[/tex]
[tex] x\in \Big(\dfrac{\pi}{2},\pi\Big) \Rightarrow \sin \Big(\dfrac{x}{2}\Big) =+\sqrt{\dfrac{9}{10}} \Rightarrow \sin \Big(\dfrac{x}{2}\Big) = \dfrac{3}{\sqrt{10}} \Rightarrow \\ \\ \Rightarrow \boxed{\sin \Big(\dfrac{x}{2}\Big) = \dfrac{3\sqrt{10}}{10}}[/tex]
