Question

show that $\cos {38}^{\circ }\cos {52}^{\circ }-\sin {38}^{\circ }\sin {52}^{\circ }=0$

Answer 1

$\begin{array}{c}\cos {38}^{\circ }\cos {52}^{\circ }-\sin {38}^{\circ }\sin {52}^{\circ }\\ =\cos \left({90}^{\circ }-{52}^{\circ }\right)\cos \left({90}^{\circ }-{38}^{\circ }\right)-\sin {38}^{\circ }\sin {52}^{\circ }\\ =\sin {52}^{\circ }\sin {38}^{\circ }-\sin {38}^{\circ }\sin {52}^{\circ }=0\end{array}$

proved