Answer
$1$
Work Step by Step
$\sin \dfrac {5\pi }{9}\cos \dfrac {\pi }{18}-\cos \dfrac {5\pi }{9}\sin \dfrac {\pi }{18}=\sin \left( \dfrac {5\pi }{9}-\dfrac {\pi }{18}\right) =\sin \dfrac {\pi }{2}=1$
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