\(\displaystyle tan(\theta) \ >0 \ and \ sin(\theta) \ < \ 0, \ ergo \ third \ quadrant.\)
\(\displaystyle Given: \ tan(\theta) \ = \ \frac{-7}{-24} \ = \ \frac{7}{24} \ and \ sin(\theta) \ = \ \frac{-7}{25} \ < \ 0, \ find \ cos(2\theta).\)
\(\displaystyle cos(\theta) \ = \ \frac{-24}{25}, \ cos^2(\theta) \ = \ \frac{576}{625}.\)
\(\displaystyle cos^2(\theta) \ = \ \frac{1+cos(2\theta)}{2} \ = \ \frac{576}{625}.\)
\(\displaystyle cos(2\theta) \ = \ \frac{2(576)}{625}-1 \ = \ \frac{527}{625}.\)
\(\displaystyle Hence, \ 2\theta \ = \ arccos\bigg(\frac{527}{625}\bigg) \ \dot= \ 32.52^0.\)
\(\displaystyle But \ cos(2\theta) \ is \ in \ third \ quadrant, \ therefore \ 2\theta \ = \ 32.52^0+180^0 \ = \ 212.52^0.\)
\(\displaystyle Ergo, \ cos(2\theta) \ = \ cos(212.52^0).\)
