Answer: [tex]\dfrac{6\sqrt{2}+\sqrt{7}}{12},\ \dfrac{-3-2\sqrt{14}}{12}[/tex]
Step-by-step explanation:
Given
x and y is in the fourth quadrant
[tex]\sin x=-\dfrac{1}{3}\\\\\cos y=\dfrac{3}{4}[/tex]
we know,
[tex]\cos (x-y)=\cos x\cos y+\sin x\sin y\\\sin (x+y)=\sin x\cos y+\cos x\sin y[/tex]
using [tex]\sin ^2\theta +\cos^2 \theta =1[/tex]
[tex]\Rightarrow \cos x=\dfrac{2\sqrt{2}}{3}\\\\\Rightarrow \sin y=-\dfrac{\sqrt{7}}{4}[/tex]
[tex]\Rightarrow \cos(x-y)=(\dfrac{2\sqrt{2}}{3})\times (\dfrac{3}{4})+(-\dfrac{1}{3})\times (-\dfrac{\sqrt{7}}{4})=\dfrac{6\sqrt{2}+\sqrt{7}}{12}\\\\\\\Rightarrow \sin (x+y)=(-\dfrac{1}{3})\times \dfrac{3}{4}+\dfrac{2\sqrt{2}}{3}\times (\dfrac{-\sqrt{7}}{4})=\dfrac{-3-2\sqrt{14}}{12}[/tex]
