Question

Find m if $\sin {12}^0\sin {18}^0\sin {42}^0\sin {48}^0\sin {72}^0\sin {78}^0=\frac{\cos {18}^0}{2m}$.

Answer 1

$\sin {12}^{\circ }\sin {18}^{\circ }\sin {42}^{\circ }\sin {48}^{\circ }\sin {72}^{\circ }$

$=\sin {12}^{\circ }\cos {12}^{\circ }\sin {18}^{\circ }\cos {18}^{\circ }\sin {42}^{\circ }\cos {42}^{\circ }$

$=\frac{\left(2\sin {12}^{\circ }\cos {12}^{\circ }\right)\left(2\sin {18}^{\circ }\cos {18}^{\circ }\right)\left(2\sin {42}^{\circ }\cos {42}^{\circ }\right)}{8}$

$=\frac{\sin {24}^{\circ }\sin {36}^{\circ }\sin {84}^{\circ }}{8}$

$=\frac{\cos 6^{\circ }\sin (30-6{)}^{\circ }\sin (30+6{)}^{\circ }}{8}$

$=\frac{\cos 6^{\circ }\left(\frac{{\cos }^{2}6{}^{\circ }-3{\sin }^2{6}^{\circ }}{4}\right)}{8}$

$=\frac{4{\cos }^3{6}^{\circ }-3\cos 6^{\circ }}{32}$

$\Rightarrow \frac{\cos {18}^{\circ }}{2m}=\frac{\cos {18}^{\circ }}{32}$

$\therefore m=16$