Question

Aratati ca radical din 1+sinx - radical din 1-sinx = 2sin x/2, x apartine ( 0, π/2 )

Answer 1

$E = \sqrt{1-\sin x} - \sqrt{1+\sin x} \Big|^2 \\ \\ E^2 = \sqrt{1-\sin x}^2 - 2\sqrt{1-\sin x}\cdot \sqrt{1+\sin x} + \sqrt{1+\sin x}^2\\ E^2 = 1-\sin x -2\sqrt{(1-\sin x)(1+\cos x)} + 1+\sin x \\ E^2 = 1-\sin x +1 +\sin x - 2\sqrt{1^2 - \sin^2 x} \\E^2 = 2 - 2\sqrt{1-\sin^2 x} \\ E^2 = 2-2\sqrt{\cos^2 x} \\ E^2 = 2-2|\cos x|,\quad x\in \Big(0,\dfrac{\pi}{2}\Big) \Rightarrow \cos x \ \textgreater \ 0 \\ E^2 = 2-2\cos x \\ E^2 = 2(1-\cos x)$

$\sin^2 x + \cos^2 x = 1 \\ \sin^2 x = 1-\cos^2 x \\ \sin^2 x = \sin^2 x + \cos^2 x - \cos^2 x \\ \sin^2 x = -(\cos^2 x -\sin^2 x) + \cos^2 x \\ \sin^2 x = -\cos 2x + \cos^2 x -\cos^2 x - \sin^2 x + 1 \\ \sin^2 x = -\cos2x - \sin^2 x + 1 \\ \sin^2 x + \sin^2 x = 1 - \cos 2x \\ 2\sin^2 x = 1-\cos 2x \\ \sin^2 x = \dfrac{1-\cos 2x}{2} \\ \boxed{\sin^2 \Big(\frac{x}{2}\Big) = \dfrac{1- \cos x}{2}}$

$E^2= 4\cdot \dfrac{1-\cos 2x}{2} \\ \\ E^2 = 4\cdot \sin^2 \Big(\dfrac{x}{2}\Big) \\ E = \sqrt{4\cdot \sin^2 \Big(\dfrac{x}{2}\Big)} \\ E = 2\cdot \Big|\sin \Big(\dfrac{x}{2}\Big)\Big|\\ \\ x\in \Big(0,\dfrac{\pi}{2}\Big) \Rightarrow \dfrac{x}{2} \in \Big(0,\dfrac{\pi}{4}\Big) \Rightarrow \sin\Big(\dfrac{x}{2}\Big) \ \textgreater \ 0 \\ \\ \Rightarrow \boxed{\boxed{E = 2 \sin \Big(\frac{x}{2}\Big)}}_\Big {\slash \Big\slash}$