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Daca sin(a)=3/5
a ∈ ( pi/2,pi)
Cat este cos(a/2) ?
mie mi-a dat cosa=4/5
si cos(a/2) cu formula unghiului dublu - radical din ( 9/10 )
Doar ca nu este un raspuns bun si nu stiu unde e greseala


Răspuns :

2sin(a/2)cos(a/2)=3/5
notăm x =cos(a/2)
2*√(1-x^2)* x= 3/5
(1-x^2)*x^2=9/100
y=x^2
y-y^2-9/100=0
100y°2-100y+9=0
y1=[100-√(10000-4*100*9)]/200=1/10
y-2=180/200=9/10
x1=1/√10. x2=-1/√10
X3=3/√10. x-4= -3/√10
cum a/2<pi/2
sunt valabile doar valorile pozitive ( deci 2 valori!)

[tex]\it a\in\Big(\dfrac{\pi}{2},\ \pi\Big) \Rightarrow \cos a < 0\\ \\ \\ \cos a =-\sqrt{1-sin^2a} =-\sqrt{1-\dfrac{9}{25}}=-\sqrt{\dfrac{16}{25}}=-\dfrac{4}{5}\\ \\ \\ a\in\Big(\dfrac{\pi}{2},\ \pi\Big) \Rightarrow \dfrac{a}{2} \in\Big(0,\ \dfrac{\pi}{2}\Big) \Rightarrow cos\dfrac{a}{2}>0\ \ \ \ \ (*)\\ \\ \\ -\dfrac{4}{5} =\cos a = \cos 2\cdot\dfrac{a}{2} =2cos^2\dfrac{a}{2}-1 \Rightarrow cos^2\dfrac{a}{2}=\dfrac{\dfrac{1}{5}}{2} \Rightarrow cos^2\dfrac{a}{2} =\dfrac{1}{10}\stackrel{(*)}{\Longrightarrow}[/tex]

[tex]\it \Rightarrow cos\dfrac{a}{2} =\dfrac{1}{\sqrt{10}}=\dfrac{\sqrt{10}}{10}[/tex]