Korzystam ze wzorów redukcyjnych:
cos(180°-α) = sinα
tg(180°-α)= -tgα
sin(180°-α) = sinα
[tex]cos150^{0} =cos(180^{0} -30^{0} )=-cos30^{0} =-\dfrac{\sqrt{3} }{2} \\\\tg120^{0} =cos(180^{0} -60^{0} )=-tg60^{0} =-\sqrt{3} \\\\tg135^{0} =cos(180^{0} -45^{0} )=-tg45^{0}=-1\\\\sin120^{0}=sin(180^{0} -60^{0} )= \dfrac{\sqrt{3} }{2} \\\podstawiam[/tex]
podstawiam obliczone wartości:
[tex]\dfrac{ cos150^{0} \cdot tg120^{0} }{\sqrt{3} \cdot sin120^{0} -tg135^{0} } =\dfrac{( -\dfrac{\sqrt{3} }{2})\cdot ( -\sqrt{3}) }{\sqrt{3} \cdot \dfrac{\sqrt{3} }{2} - ( -1)} =\dfrac{\dfrac{3}{2} }{\dfrac{3}{2}+1 } =\dfrac{\dfrac{3}{2} }{\dfrac{5}{2} }=\dfrac{3}{2} \div \dfrac{5}{2} =\dfrac{3}{2} \cdot \dfrac{2}{5} =\dfrac{3}{5}[/tex]
