Question

sin480° cos690°+cos780°sin1050°​

Answer 1

$\begin{aligned}&\sin480\textdegree\cos690\textdegree+\cos780\textdegree\sin1050\textdegree\\\\\implies\ \ &\sin(360\textdegree+120\textdegree)\cos(720\textdegree-30\textdegree)+\cos(720\textdegree+60\textdegree)\sin(1080\textdegree-30\textdegree)\\\\\implies\ \ &\sin(180\textdegree-60\textdegree)\cos(-30\textdegree)+\cos60\textdegree\sin(-30\textdegree)\\\\\implies\ \ &\sin60\textdegree\cos30\textdegree-\cos60\textdegree\sin30\textdegree\end{aligned}$

$\\\\\implies\ \ &\sin(60\textdegree-30\textdegree)\ =\ \sin30\textdegree=\mathbf{\dfrac{1}{2}}$