Question

Ştiind că x ∈ (π/2, π) şi ca sin x=3/5, să se calculeze sin ×/2.

Îmi puteți explica și rezolvarea pas cu pas va rog? ❤❤❤

Answer 1

Stim formula:
sin2a=2sina·cosa  
Fie 2a=x
sinx=2sinx/2 · cosx/2 
Dar sin²x/2 + cos²x/2=1 (teorema fundamentala a trigonometriei)
=> cosx/2=√(1-sin²x/2)
Inlocuim si obtinem
2sinx/2·√(1-sin²x/2)=3/5
Notam  sinx/2=y unde y∈(-1 ,1)
2y·√(1-y²)=3/5   l²
4y²(1-y²)=9/25
Facem  substitutia  y²=t unde t∈(0,1) si ecuatia devine
4t(1-t)=9/25
4t²-4t-9/25=0  l·25
100t²-100t+9=0
t₁=1/10=>y²=1/10=>y₁=1/√10=√10/10 (s-a rationalizat  numitorul) si
                                y₂=-√10/10=>sinx/2=√10/10
t₂=9/10=>y=3/√10=3√10/10
sinx/2=3√10/10
(S-au luat numai valorile pozitive, fiindca x/2∈(π/4 ,π/2), interval pe care sinusul este pozitiv.)

Answer 2

$\\ Il scoatem pe \sin\Big(\dfrac{x}{2}\Big).\\ \\ \cos2x = \cos^2 x-\sin^2 x\Rightarrow \cos2x = (1-\sin^2 x)-\sin^2 x \Rightarrow \\ \\ \Rightarrow \cos2x = 1-\sin^2 x-\sin^2 x \Rightarrow \cos2x = 1-2\sin^2 x \Rightarrow \\ \\ \Rightarrow 2\sin^2 x = 1 - \cos2x \Rightarrow \sin^2 x = \dfrac{1-\cos2x}{2} \Rightarrow \\ \\ \Rightarrow \sin^2 \Big(\dfrac{x}{2}\Big) = \dfrac{1-\cos \Big(2\cdot \dfrac{x}{2}\Big)}{2} \Rightarrow \sin^2 \Big(\dfrac{x}{2}\Big) = \dfrac{1-\cos x}{2}$

$\\ Avem nevoie de \cos x.\\ \\ \sin^2x+\cos^2x = 1 \Rightarrow \cos^2x = 1-\sin^2x \Rightarrow \cos^2 x = 1-\Big(\dfrac{3}{5}\Big)^2 \Rightarrow \\ \\ \Rightarrow \cos^2x = 1 - \dfrac{9}{25} \Rightarrow \cos^2 x = \dfrac{25-9}{25} \Rightarrow \cos^2 x = \dfrac{16}{25} \Rightarrow \\ \\ \Rightarrow\cos x = \pm \sqrt{\dfrac{16}{25}} \Rightarrow \cos x = \dfrac{4}{5}, dar, x\in \Big(\dfrac{\pi}{2},\pi\Big) \Rightarrow \cos x = -\dfrac{4}{5}$

$\sin^2 \Big(\dfrac{x}{2}\Big) = \dfrac{1-\Big(-\dfrac{4}{5}\Big)}{2} \Rightarrow \sin^2 \Big(\dfrac{x}{2}\Big) =\dfrac{1+\dfrac{4}{5}}{2} \Rightarrow \sin^2 \Big(\dfrac{x}{2}\Big) =\dfrac{\dfrac{5+4}{5}}{2} \Rightarrow \\ \\ \Rightarrow \sin^2 \Big(\dfrac{x}{2}\Big) =\dfrac{5+4}{2\cdot 5} \Rightarrow \sin^2 \Big(\dfrac{x}{2}\Big) =\dfrac{9}{10} \Rightarrow \sin \Big(\dfrac{x}{2}\Big) =\pm\sqrt{\dfrac{9}{10}} , dar,$
$x\in \Big(\dfrac{\pi}{2},\pi\Big) \Rightarrow \sin \Big(\dfrac{x}{2}\Big) =+\sqrt{\dfrac{9}{10}} \Rightarrow \sin \Big(\dfrac{x}{2}\Big) = \dfrac{3}{\sqrt{10}} \Rightarrow \\ \\ \Rightarrow \boxed{\sin \Big(\dfrac{x}{2}\Big) = \dfrac{3\sqrt{10}}{10}}$