Given:
[tex]\sin (x)=-\dfrac{1}{2}[/tex]
[tex]\cos (y)=-\dfrac{1}{3}[/tex]
x is an angle in quadrant III, and y is an angle in quadrant II.
To find:
The value of sin(x+y).
Solution:
We know that, only sin and cosec are positive in II quadrant. Only tan and cot are positive in III quadrant.
x is an angle in quadrant III,
[tex]\cos x=-\sqrt{1-\sin^2 x}[/tex]
[tex]\cos x=-\sqrt{1-(-\dfrac{1}{2})^2}[/tex]
[tex]\cos x=-\sqrt{1-\dfrac{1}{4}}[/tex]
[tex]\cos x=-\sqrt{\dfrac{4-1}{4}}[/tex]
[tex]\cos x=-\dfrac{\sqrt{3}}{2}[/tex]
y is an angle in quadrant II.
[tex]\sin y=\sqrt{1-\cos^2x}[/tex]
[tex]\sin y=\sqrt{1-(-\dfrac{1}{3})^2}[/tex]
[tex]\sin y=\sqrt{1-\dfrac{1}{9}}[/tex]
[tex]\sin y=\sqrt{\dfrac{9-1}{9}}[/tex]
[tex]\sin y=\sqrt{\dfrac{8}{9}}[/tex]
[tex]\sin y=\dfrac{2\sqrt{2}}{3}[/tex]
Now,
[tex]\sin(x+y)=\sin x\cos y+\cos x\sin y[/tex]
[tex]\sin(x+y)=(-\dfrac{1}{2})\times (-\dfrac{1}{3})+(-\dfrac{\sqrt{3}}{2})\times (\dfrac{2\sqrt{2}}{3})[/tex]
[tex]\sin(x+y)=\dfrac{1}{6}-\dfrac{2\sqrt{6}}{6}[/tex]
[tex]\sin(x+y)=\dfrac{1-2\sqrt{6}}{6}[/tex]
Therefore, the correct option is B.
