Answer:
[tex](a)-\dfrac{11\sqrt{5}}{25} \\(b) -\dfrac{2\sqrt{5}}{25} \\(c)\dfrac{11}{2}[/tex]
Step-by-step explanation:
[tex]\tan \alpha =\dfrac12, \pi < \alpha< \dfrac{3 \pi}{2}[/tex]
Therefore:
[tex]\alpha$ is in Quadrant III[/tex]
Opposite = -1
Adjacent =-2
Using Pythagoras Theorem
[tex]Hypotenuse^2=Opposite^2+Adjacent^2\\=(-1)^2+(-2)^2=5\\Hypotenuse=\sqrt{5}[/tex]
Therefore:
[tex]\sin \alpha =-\dfrac{1}{\sqrt{5}}\\\cos \alpha =-\dfrac{2}{\sqrt{5}}[/tex]
Similarly
[tex]\cos \beta =\dfrac35, \dfrac{3 \pi}{2}<\beta<2\pi\\\beta $ is in Quadrant IV (x is negative, y is positive), therefore:\\Adjacent=$-3\\$Hypotenuse=5\\Opposite=4 (Using Pythagoras Theorem)[/tex]
[tex]\sin \beta =\dfrac{4}{5}\\\tan \beta =-\dfrac{4}{3}[/tex]
(a)
[tex]\cos(\alpha + \beta)=\cos\alpha\cos\beta-\sin \alpha\sin \beta\\[/tex]
[tex]=-\dfrac{2}{\sqrt{5}}\cdot \dfrac{3}{5}-(-\dfrac{1}{\sqrt{5}})(\dfrac{4}{5})\\=-\dfrac{2\sqrt{5}}{5}\cdot \dfrac{3}{5}+\dfrac{\sqrt{5}}{5}\cdot\dfrac{4}{5}\\=-\dfrac{2\sqrt{5}}{25}[/tex]
(b)
[tex]\sin(\alpha + \beta)=\sin\alpha\cos\beta+\cos \alpha\sin \beta[/tex]
[tex]\sin(\alpha + \beta)=\sin\alpha\cos\beta+\cos \alpha\sin \beta\\=-\dfrac{1}{\sqrt{5}}\cdot\dfrac35+(-\dfrac{2}{\sqrt{5}})(\dfrac{4}{5})\\=-\dfrac{\sqrt{5}}{5}\cdot\dfrac35-\dfrac{2\sqrt{5}}{5}\cdot\dfrac{4}{5}\\=-\dfrac{11\sqrt{5}}{25}[/tex]
(c)
[tex]\tan(\alpha + \beta)=\dfrac{\sin(\alpha + \beta)}{\sin(\alpha + \beta)}=-\dfrac{11\sqrt{5}}{25} \div -\dfrac{2\sqrt{5}}{25} =\dfrac{11}{2}[/tex]