Hi Sarcasticcarma!
\(cos(\theta)=-\dfrac{\sqrt{2}}{6} \text{ where } \pi \le \theta \le \dfrac{3\pi}{2}\)
\(tan(\beta)=\dfrac{5}{12} \text{ where } 0 \le \beta \le \dfrac{\pi}{2}\)
\(\text{What is the exact value of } sin(\theta+\beta)?\)
For this question, we need to know the expansion of \(sin(A+B)\) in general.
So remember: \(sin(A+B)=sin(A)cos(B)+sin(B)cos(A)\)
So: \(sin(\theta+\beta)=sin(\theta)cos(\beta)+sin(\beta)cos(\theta)\) (1)
To solve this question, we just need to find: \(sin(\theta),cos(\beta),sin(\beta),cos(\theta)\) (But we already have \(cos(\theta)\)).
If \(cos(\theta)=-\dfrac{\sqrt{2}}{6}\), then we have two ways to get \(sin(\theta)\).
First way: Recall: \(cos^2(\theta)+sin^2(\theta)=1 \iff sin(\theta)=\pm \sqrt{1-cos^2(\theta)}\)
So: \(sin(\theta)=\pm \sqrt{1-\dfrac{{2}}{36}}=\pm\sqrt{\dfrac{17}{18}} \\ \) But shall we take the positive or negative?
We must look at the given interval of \(\theta\) which is between \(\pi \text{ and } \dfrac{3\pi}{2}\)
This is the third quadrant, so \(sin(\theta)\) is negative. (Determined by the "CAST" rule or drawing sine graph.)
Thus, \(sin(\theta)=-\sqrt{\dfrac{17}{18}}\)
The second way to get \(sin(\theta)\) is much faster and in fact, we will use it to get \(cos(\beta),sin(\beta)\), as follows:
Given: \(tan(\beta)=\dfrac{5}{12}\)
Then, draw a right angle triangle, and choose any angle to be \(\beta\)
Then, we know \(tan(\beta)=\dfrac{5}{12}\), so the opposite side to the angle you chose is 5 and the adjacent must be 12; and by pythagoras theorem, the hypotenuse is 13. (Draw it!)
Now what is \(sin(\beta)?,cos(\beta)?\) This is easily done via the triangle we already drew!
So:
\(sin(\beta)=\dfrac{5}{13}\\ \\ \\ cos(\beta)=\dfrac{12}{13}\)
And since we are in the first quadrant, all of them will be positive.
So finally by (1):
\(sin(\theta+\beta)=sin(\theta)cos(\beta)+sin(\beta)cos(\theta) \\ sin(\theta+\beta)=-\sqrt{\dfrac{17}{18}}*\dfrac{12}{13}+\dfrac{5}{13}*(-\dfrac{\sqrt{2}}{6})\)
You can simplify it :).
I hope this helps!
!
