Answer: [tex]\sin(x+y)=\dfrac{1-8\sqrt{3}}{15}[/tex]
Step-by-step explanation:
Since we have given that
[tex]\sin x=\dfrac{1}{3}\\\\so,\\\\\cos x=\sqrt{1-\dfrac{1}{9}}=\sqrt{\dfrac{8}{9}}=\dfrac{2\sqrt{2}}{3}[/tex]
Since x ends in the 2 nd quadrant,
So, [tex]\cos x=\dfrac{-2\sqrt{2}}{3}[/tex]
Similarly,
[tex]\cos y=\dfrac{1}{5}\\\\So,\\\\\sin y=\sqrt{1-\dfrac{1}{25}}=\sqrt{\dfrac{24}{25}}=\dfrac{2\sqrt{6}}{5}[/tex]
So, sin(x+y) is given by
[tex]\sin x\cos y+\sin y\cos x\\\\\\=\dfrac{1}{3}\times \dfrac{1}{5}+\dfrac{2\sqrt{6}}{5}\times (-)\dfrac{2\sqrt{2}}{3}\\\\\\=\dfrac{1}{15}-\dfrac{8\sqrt{3}}{15}\\\\\\=\dfrac{1-8\sqrt{3}}{15}[/tex]
Hence, [tex]\sin(x+y)=\dfrac{1-8\sqrt{3}}{15}[/tex]
