Recall that \(\sin^2x+\cos^2x=1\).
This means
\(\cos u=\sqrt{1-\sin^2u}=\sqrt{1-\dfrac19}=\dfrac{\sqrt8}3=\dfrac{2\sqrt2}3\)
\(cos v=sqrt{1-\sin^2v}=\sqrt{1-\dfrac1{16}}=\dfrac{\sqrt{15}}4\)
\(\sin(u+v)=\sin u\cos v+\sin v\cos u=\dfrac13\times\dfrac{\sqrt{15}}4+\dfrac14\times\dfrac{2\sqrt2}3=\dfrac{2\sqrt2+\sqrt{15}}{12}\)
