Answer:
[tex]\large\boxed{Q2.\ \cos\theta=\dfrac{3}{5}}\\\boxed{Q3.\ \sin(\alpha+\beta)=-\dfrac{13}{85}}[/tex]
Step-by-step explanation:
Q2.
[tex]270^o<\theta<360^o-IV\ Quarter\\\\\sin\theta=-\dfrac{4}{5},\ \cos\theta=?\\\\\text{Use}\ \sin^2x+\cos^2x=1\\\\\left(-\dfrac{4}{5}\right)^2+\cos^2\theta=1\\\\\dfrac{16}{25}+\cos^2\theta=1\qquad\text{subtract}\ \dfrac{16}{25}\ \text{from both sides}\\\\\cos^2\theta=\dfrac{25}{25}-\dfrac{16}{25}\\\\\cos^2\theta=\dfrac{9}{25}\to\cos\theta=\pm\sqrt{\dfrac{9}{25}}\\\\\cos\theta=\pm\dfrac{3}{5}\\\\\theta\ \text{is in IV Quarter. Therefore}\ \cos\theta>0\to\boxed{\cos\theta=\dfrac{3}{5}}[/tex]
Q3.
[tex]180^o\alpha<270^o-III\ Quarter\\\\\cos\alpha=-\dfrac{8}{17},\\\\\sin^2\alpha+\left(-\dfrac{8}{17}\right)^2=1\\\\\sin^2\alpha+\dfrac{64}{289}=1\qquad\text{subtract}\ \dfrac{64}{289}\ \text{from both sides}\\\\\sin^2\alpha=\dfrac{289}{289}-\dfrac{64}{289}\\\\\sin^2\alpha=\dfrac{225}{289}\to\sin\alpha=\pm\sqrt{\dfrac{225}{289}}\\\\\sin\alpha=\pm\dfrac{15}{17}\\\\\alpha\ \text{is in III Quarter. Therefore}\ \sin\alpha=-\dfrac{15}{17}[/tex]
[tex]270^o<\beta<360^o-IV\ Quarter\\\\\sin\beta=-\dfrac{4}{5}\\\\\left(-\dfrac{4}{5}\right)^2+\cos^2\beta=1\\\\\dfrac{16}{25}+\cos^2\beta=1\qquad\text{subtract}\ \dfrac{16}{25}\ \text{from both sides}\\\\\cos^2\beta=\dfrac{25}{25}-\dfrac{16}{25}\\\\\cos^2\beta=\dfrac{9}{25}\to\cos\beta=\pm\sqrt{\dfrac{9}{25}}\\\\\cos\beta=\pm\dfrac{3}{5}\\\\\beta\ \text{is in IV Quarter. Therefore}\ \cos\beta=\dfrac{3}{5}[/tex]
[tex]\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha\\\\\text{Substitute:}\\\\\sin(\alpha+\beta)=\left(-\dfrac{15}{17}\right)\left(\dfrac{3}{5}\right)+\left(-\dfrac{4}{5}\right)\left(-\dfrac{8}{17}\right)\\\\=-\dfrac{45}{85}+\dfrac{32}{85}=-\dfrac{13}{85}[/tex]
