Question

Aratati ca √(1+sinx) - √(1-sin x)= 2 sin (x/2)
x∈(0, π/2)

Answer 1

daca x apartine intervalului [0,90] grade, atunci sinx>0 si cosx>0Ridicam ecuatia la patrat$(\sqrt{1+sinx}-\sqrt{1-sinx})^{2}=\sqrt{1+sinx}^{2}+\sqrt{1-sinx}^{2}-2\sqrt{(1+sinx)(1-sinx)}=1+sinx+1-sinx+\sqrt{1-\sin{x}^{2}}=2-2\sqrt{\cos{x}^{2}}=2-2cosx=4\sin{\frac{x}{2}}^{2}\Rightarrow 1-cosx=2\sin{\frac{x}{2}}^{2}$
Stim relatia general valabila
$\cos{{2x}}=1-2\sin{x}^{2}$ daca inlocuim in formula 2x cu x, obtinem atunci$\cos{x}=1-2\sin{\frac{x}{2}}^{2}\Rightarrow 2\sin{\frac{x}{2}}^{2}=1-\cos{x}$

care este fix relatia mai sus de demonstrat.

Answer 2

[tex]\it \sqrt{1+sinx} -\sqrt{1-sinx} =k \Rightarrow (\sqrt{1+sinx} -\sqrt{1-sinx})^2 =k^2 \Rightarrow \\ \\ \\ \Rightarrow 1+sinx-2\sqrt{(1+sinx)(1-sinx)} +1-sinx = k^2\Rightarrow \\ \\ \\ \Rightarrow 2-2\sqrt{1-sin^2x} = k^2 \Rightarrow 2-2cosx = k^2\Rightarrow 2-2cos2\cdot\dfrac{x}{2} =k^2 \Rightarrow [/tex]

$\it \Rightarrow 2-2(1-2sin^2\dfrac{x}{2}) = k^2 \Rightarrow 2-2+4sin^2\dfrac{x}{2} =k^2 \Rightarrow 4sin^2\dfrac{x}{2} = k^2\Rightarrow\\ \\ \\ \Rightarrow\sqrt{4sin^2\dfrac{x}{2}} = \sqrt{k^2} \Rightarrow 2sin\dfrac{x}{2} = k \Rightarrow k = 2sin\dfrac{x}{2}$


Observație:

$\it x \in \left(0,\dfrac{\pi}{2} \right) \Rightarrow \sin x \ \textgreater \ 0,\ \ \cos x\ \textgreater \ 0.$