Question

Show that $\sin (-{420}^{\circ })\cos \left({390}^{\circ }\right)+\cos (-{660}^{\circ })\sin \left({330}^{\circ }\right)=-1$.

Answer 1

$\sin \left({-420}^0\right)\cos \left({390}^0\right)+\cos \left({-660}^0\right)\sin \left({330}^0\right)=-\sin \left({420}^0\right)\cos \left({390}^0\right)\cos \left({660}^0\right)\sin \left({330}^0\right)=-\sin ({360}^0+{120}^0)\cos ({360}^0+{330}^0)+\cos ({360}^0+{300}^0)\sin \left({360}^0{30}^0\right)=-\sin \left({120}^0\right)\cos \left({30}^0\right)-\cos \left({360}^0\right)\sin \left({-30}^0\right)=-\sin ({180}^0+{60}^0)\cos \left({30}^0\right)-\cos ({360}^0-{60}^0)\sin {30}^0=-\sin {60}^0\cos {30}^0-\cos {60}^0\sin {30}^0=-\sin ({30}^0+{60}^0)=-\sin {90}^0=-1$