cosθ=−√2/6 , where π≤θ≤3π/2 .

Thanks Asinus,

Also asked and answered here

https://web2.0calc.com/questions/please-help-me-i-m-having-a-hard-time-understanding

What is the exact value of sin(θ+β)?

Hello Sarcasticcarma! 

\(cos\ θ=−\frac{\sqrt{2}}{6}\ |\ π\le θ\le\frac{3\pi}{2}\\ tan\ β=\frac{5}{12}\ |\ 0\le\beta\le \frac{\pi}{2}\\\)

\(sin(\theta+\beta)=sin \theta\ cos\beta+cos\theta\ sin\beta\\ sin\theta=-\sqrt{1-cos^2\theta}\ \Leftarrow\ 3rd\ quadrant\\ sin\beta=\frac{tan\beta}{\sqrt{1+tan^2\beta}}\ \\ cos\beta=\frac{1}{\sqrt{1+tan^2\beta}}\\ sin(\theta+\beta)=-\sqrt{1-cos^2\theta}\cdot \frac{1}{\sqrt{1+tan^2\beta}}+cos\theta\cdot \frac{tan\beta}{\sqrt{1+tan^2\beta}}\\ \)

\(sin(\theta+\beta)=-\sqrt{1-(-\frac{\sqrt{2}}{6})^2}\cdot \dfrac{1}{\sqrt{1+(\frac{5}{12})^2}}+(-\frac{\sqrt{2}}{6})\cdot \dfrac{(\frac{5}{12})}{\sqrt{1+(\frac{5}{12})^2}}\\ \)

\(sin(\theta+\beta)=-\frac{\sqrt{34}}{6}\ \cdot \ \frac{12}{13}\ -\ \frac{\sqrt{2}}{6}\ \cdot \ \frac{5}{ 13}\\ \color{blue}sin(\theta+\beta)=-\dfrac{12\cdot \sqrt{34}+5\cdot \sqrt{2}}{78}=-0.9877\)

  !