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Sagot :
Given
sin x = − 2/3 and x is in Quadrant III; and sin y = 1/3 and y is in Quadrant II.
Find
sin(x + y)
Explanation
sin x = − 2/3 and x is in Quadrant III
as, cos x is negative in Quadrant III
[tex]\begin{gathered} \cos x=-\sqrt{1-\sin^2x} \\ \\ \cos x=-\sqrt{1-(-\frac{2}{3})^2} \\ \\ \cos x=-\sqrt{1-\frac{4}{9}} \\ \\ \cos x=-\sqrt{\frac{5}{9}} \\ \\ \cos x=\frac{-\sqrt{5}}{3} \end{gathered}[/tex]sin y = 1/3 and y is in Quadrant II.
cos y is negative in Quadrant II.
[tex]\begin{gathered} \cos y=-\sqrt{1-\sin^2xy} \\ \\ \cos y=-\sqrt{1-(\frac{1}{3})^2} \\ \\ \cos y=-\sqrt{1-\frac{1}{9}} \\ \\ \cos y=-\sqrt{\frac{8}{9}} \\ \\ \cos y=\frac{-2\sqrt{2}}{3} \end{gathered}[/tex]as we know the formula of
sin (x + y)= sinx cosy + cosx siny
now put values ,
[tex]\begin{gathered} \sin(x+y)=\sin x\cos y+\cos x\sin y \\ \\ \sin(x+y)=(-\frac{2}{3})(\frac{-2\sqrt{2}}{3})+(\frac{-\sqrt{5}}{3})(\frac{1}{3}) \\ \\ \sin(x+y)=\frac{4\sqrt{2}}{9}-\frac{\sqrt{5}}{9} \\ \\ \sin(x+y)=\frac{4\sqrt{2}-\sqrt{5}}{9} \end{gathered}[/tex]Final Answer
[tex]\sin(x+y)=\frac{4\sqrt{2}-\sqrt{5}}{9}[/tex]
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