[tex]f\left(\dfrac\beta2\right)=\sin\dfrac\beta2[/tex]
Recall the half-angle identity:
[tex]\sin^2\dfrac\beta2=\dfrac{1-\cos\beta}2\implies\sin\dfrac\beta2=\pm\sqrt{\dfrac{1-\cos\beta}2}[/tex]
So in order to find [tex]f\left(\dfrac\beta2\right)[/tex], we only need to know the value of [tex]\cos\beta[/tex]. But there are two possible values. To decide which to take, we use the given information: we know that [tex]\beta[/tex] lies in quadrant III, which means [tex]\pi<\beta<\dfrac{3\pi}2[/tex], so [tex]\dfrac\pi2<\dfrac\beta2<\dfrac{3\pi}4[/tex], which places [tex]\dfrac\beta2[/tex] in quadrant II. In this quadrant, the sine of angle is positive, and so
[tex]\sin\dfrac\beta2=\sqrt{\dfrac{1-\cos\beta}2}[/tex]
[tex]\beta[/tex] is an angle of a ray whose terminal point, [tex]\left(-\dfrac13,y\right)[/tex] lies on the circle [tex]x^2+y^2=1[/tex]. The [tex]x[/tex]-coordinate is all we need to know in order to find that [tex]\cos\beta=-\dfrac13[/tex]. So,
[tex]\sin\dfrac\beta2=\sqrt{\dfrac{1-\left(-\frac13\right)}2}=\sqrt{\dfrac46}=\dfrac{\sqrt2}{\sqrt3}=\dfrac{\sqrt6}3[/tex]
