Question

Let theta be an angle between pi and 3pi/2 such that cos(theta) = -1/9

Answer 1

so we know that θ is between π and 3π/2, which is another of saying that θ is in the III Quadrant, and obviously its cosine will be negative.

we also know that cos(θ/2) is negative, well, that rules out the I and IV Quadrants, so most likely θ/2 is on the II Quadrant, since is smaller than θ anyway, and on the II Quadrant as we know, the sine or "y" value is positive.

`\stackrel{\textit{Half-Angle Identities}}{ sin\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-cos(\theta)}{2}} \qquad\qquad cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}}} \\\\[-0.35em] ~\dotfill\\\\ cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+\left(-(1)/(9) \right)}{2}}\implies cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-\left((1)/(9) \right)}{2}}`

`cos\left(\cfrac{\theta}{2}\right)=-\sqrt{\cfrac{~~ (8)/(9) ~~}{2}}\implies cos\left(\cfrac{\theta}{2}\right)=-\sqrt{\cfrac{4}{9}}\implies cos\left(\cfrac{\theta}{2}\right)=-\cfrac{2}{3} \\\\[-0.35em] ~\dotfill`

`sin\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-\left( -(1)/(9) \right)}{2}}\implies sin\left(\cfrac{\theta}{2}\right)=+ \sqrt{\cfrac{~~(10)/(9)~~}{2}}\implies sin\left(\cfrac{\theta}{2}\right)=+\cfrac{√(5)}{3} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{~~~a_1~\hfill a_2~~~}{\left( -\cfrac{2}{3}~~,~~ \cfrac{√(5)}{3}\right)}~\hfill`

Answer 2

(-2/3, (√5)/3)

Explanation:

We can use the half-angle identities to find the values of the sine and cosine of the desired angle. The desired angle is a 2nd-quadrant angle, so the sine is positive, and the cosine is negative.

`\sin{\left((\theta)/(2)\right)}=\sqrt{(1-cos((\theta)))/(2)}=\sqrt{(1-(-1/9))/(2)}=\sqrt{(5)/(9)}=(√(5))/(3)\\\\\cos{\left((\theta)/(2)\right)}=-\sqrt{(1+cos((\theta)))/(2)}=-\sqrt{(1+(-1/9))/(2)}=-\sqrt{(4)/(9)}=-(2)/(3)`

The coordinates of the terminal point are ...

`(a_1,a_2)=\left(\cos{(\theta)/(2)},\sin{(\theta)/(2)}\right)=\left(-(2)/(3),(√(5))/(3)\right)`