Question

Please Help!!!

Answer 1

$f\left(\dfrac\beta2\right)=\sin\dfrac\beta2$

Recall the half-angle identity:

$\sin^2\dfrac\beta2=\dfrac{1-\cos\beta}2\implies\sin\dfrac\beta2=\pm\sqrt{\dfrac{1-\cos\beta}2}$

So in order to find $f\left(\dfrac\beta2\right)$, we only need to know the value of $\cos\beta$. But there are two possible values. To decide which to take, we use the given information: we know that $\beta$ lies in quadrant III, which means $\pi$, so $\dfrac\pi2$, which places $\dfrac\beta2$ in quadrant II. In this quadrant, the sine of angle is positive, and so

$\sin\dfrac\beta2=\sqrt{\dfrac{1-\cos\beta}2}$

$\beta$ is an angle of a ray whose terminal point, $\left(-\dfrac13,y\right)$ lies on the circle $x^2+y^2=1$. The $x$-coordinate is all we need to know in order to find that $\cos\beta=-\dfrac13$. So,

$\sin\dfrac\beta2=\sqrt{\dfrac{1-\left(-\frac13\right)}2}=\sqrt{\dfrac46}=\dfrac{\sqrt2}{\sqrt3}=\dfrac{\sqrt6}3$