a∈(π /2, π); cos a= - 1/3; sin 2a=?
sina=+√(1-cos²a)=√(1-1/9)= √(8/9=2√2 /3; am ales +; pt ca a∈(π /2, π)
sin 2a=2sina•cosa=2•2√2 /3 •(-1/3)= -4√2 /9
Răspuns:
sin(2a)=0
Explicație pas cu pas:
[tex]a=[\frac{\pi }{2} ,\pi ]\\cos(a)=(-\frac{1}{3}) \\sin(a)=?\\\\1) sin(2\pi )= ? = 0 \\= 2cos(\pi )sin(\pi )\\ cos(\pi )=(-1)\\sin(\pi )=0\\=2(-1)*0=\\(-2)*0=0\\\\2) sin(2*\frac{\pi }{2} )= ? = 0\\cos(a)=(-\frac{1}{3} )\\sin(2a)= ?\\cos(\frac{\pi }{2}) =(-\frac{1}{3} )\\sin(2*\frac{\pi }{2} )= ?\\a*\frac{b}{c} =\frac{a*b}{c} \\sin(2*\frac{\pi }{2} )=\\sin(\frac{2\pi }{2})=\\ sin(\pi )=0[/tex]
Succes! :)
