homework help

\(\pi<\theta<\frac{3\pi}{2}\qquad 3rd\;\; quad\\ \frac{\pi}{2}<\frac{\theta}{2}<\frac{3\pi}{24}\qquad 2nd\; quad\\\)

I didn't have to do this. 

You are told that cos (theta) and cos(theta/2) are negative, you didn't need to be told this.  It is just a fact.

Cos in the 2nd and 3rd quadrant will always be negative.

You are told that  cos(theta)=-1/9 and you need to find   cos(theta/2)

\(cos(\theta)=cos^2(\frac{\theta}{2})-sin^2(\frac{\theta}{2})\\ cos(\theta)=cos^2(\frac{\theta}{2})-(1-cos^2(\frac{\theta}{2}))\\ cos(\theta)=2cos^2(\frac{\theta}{2})-1\\ \sqrt{\frac{cos(\theta)+1}{2}}=cos(\frac{\theta}{2})\\ cos(\frac{\theta}{2})=\sqrt{\frac{cos(\theta)+1}{2}}\\ cos(\frac{\theta}{2})=\sqrt{\frac{\frac{-1}{9}+1}{2}}\\ cos(\frac{\theta}{2})=\sqrt{\frac{\frac{8}{9}}{2}}\\ cos(\frac{\theta}{2})=\sqrt{{\frac{8}{9*2}}}\\ cos(\frac{\theta}{2})=\pm\sqrt{{\frac{4}{9}}}\\ But \;it\;is\;neg\\ cos(\frac{\theta}{2})=-\frac{2}{3}\\ sin(\frac{\theta}{2})=\sqrt{1-\frac{4}{9}}\\ sin(\frac{\theta}{2})=\frac{\sqrt5}{3}\\ \text{The terminal point is } \quad (\frac{-2}{3},\frac{\sqrt5}{3}) \)

As always, you will need to check my working.

LaTex:

cos(\theta)=cos^2(\frac{\theta}{2})-sin^2(\frac{\theta}{2})\\
cos(\theta)=cos^2(\frac{\theta}{2})-(1-cos^2(\frac{\theta}{2}))\\
cos(\theta)=2cos^2(\frac{\theta}{2})-1\\
\sqrt{\frac{cos(\theta)+1}{2}}=cos(\frac{\theta}{2})\\
cos(\frac{\theta}{2})=\sqrt{\frac{cos(\theta)+1}{2}}\\
cos(\frac{\theta}{2})=\sqrt{\frac{\frac{-1}{9}+1}{2}}\\
cos(\frac{\theta}{2})=\sqrt{\frac{\frac{8}{9}}{2}}\\
cos(\frac{\theta}{2})=\sqrt{{\frac{8}{9*2}}}\\
cos(\frac{\theta}{2})=\pm\sqrt{{\frac{4}{9}}}\\
But \;it\;is\;neg\\
cos(\frac{\theta}{2})=-\frac{2}{3}\\
sin(\frac{\theta}{2})=\sqrt{1-\frac{4}{9}}\\
sin(\frac{\theta}{2})=\frac{\sqrt5}{3}\\
\text{The terminal point is } \quad  (\frac{-2}{3},\frac{\sqrt5}{3})

ok, thanks for the help.