Question

Prove that $\cos A\cos \cos 2A\cos 4A\cos 8A=\frac{\sin 2^{4}A}{2^4\sin A}$
Four can be used as a formula as given in and hence prove the following :
$\cos 7{}^{\circ }\cos 14{}^{\circ }\cos 28{}^{\circ }\cos 56{}^{\circ }=\frac{\sin 68{}^{\circ }}{16\cos 83{}^{\circ }}$

Answer 1

$\frac{2.\sin A\cos A.\cos 2A.\cos 4A.\cos 8A}{2.\sin A}$

$=\frac{2\sin 2A.\cos 2A.\cos 4A.\cos 8A}{2^2\sin A}$

$=\frac{2\sin 4A.\cos 4A.\cos 8A}{2^3\sin A}$

$=\frac{2\sin 8A.\cos 8A}{2^4\sin A}=\frac{\sin 16A}{2^4\sin A}=\frac{\sin 2^{4}A}{2^4\sin A}$

$\therefore \cos {57}^{\circ }.\cos {14}^{\circ }.\cos {28}^{\circ }.\cos {56}^{\circ }=\frac{\sin 16\times 7}{16\cos {83}^{\circ }}$

$=\frac{\sin {112}^{\circ }}{16\cos 83}=\frac{\sin {68}^{\circ }}{16\cos {83}^{\circ }}$