Question

Please Help!!!

So, I figured out how to find the sine and cosine of ββ, but I wasn't sure how to complete the problem. Do I find the sine of ββ and the sine of 2 separately then divide them? Is it more or less complicated than that?

Thanks for the help!

Answer 1

`f\left(\frac\beta2\right)=\sin\frac\beta2`

Recall the half-angle identity:


`\sin^2\frac\beta2=\frac{1-\cos\beta}2\implies\sin\frac\beta2=\pm\sqrt{\frac{1-\cos\beta}2}`

So in order to find
`f\left(\frac\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<\beta<\frac{3\pi}2`, so
`\frac\pi2<\frac\beta2<\frac{3\pi}4`, which places
`\frac\beta2` in quadrant II. In this quadrant, the sine of angle is positive, and so


`\sin\frac\beta2=\sqrt{\frac{1-\cos\beta}2}`


`\beta` is an angle of a ray whose terminal point,
`\left(-\frac13,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=-\frac13`. So,


`\sin\frac\beta2=\sqrt{\frac{1-\left(-\frac13\right)}2}=√(\frac46)=(\sqrt2)/(\sqrt3)=\frac{\sqrt6}3`