Discover how IDNLearn.com can help you learn and grow with its extensive Q&A platform. Ask your questions and receive prompt, detailed answers from our experienced and knowledgeable community members.
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]
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. IDNLearn.com is committed to providing accurate answers. Thanks for stopping by, and see you next time for more solutions.
