Problem Statement
Prove that:
$$\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \frac{1}{16}$$
Verified Solution & Marking Scheme
Substitute known exact value cos 60° = 1/2
$$\text{LHS} = \frac{1}{2} [\cos 20^\circ \cos 40^\circ \cos 80^\circ] = \frac{1}{4} [2 \cos 40^\circ \cos 20^\circ] \cos 80^\circ$$
Use 2 cos A cos B = cos(A+B) + cos(A-B)
$$2 \cos 40^\circ \cos 20^\circ = \cos(40^\circ + 20^\circ) + \cos(40^\circ - 20^\circ) = \cos 60^\circ + \cos 20^\circ = \frac{1}{2} + \cos 20^\circ$$
$$\text{LHS} = \frac{1}{4} \left( \frac{1}{2} + \cos 20^\circ \right) \cos 80^\circ = \frac{1}{8}\cos 80^\circ + \frac{1}{4}\cos 20^\circ \cos 80^\circ$$
Expand second term using 2 cos A cos B
$$\frac{1}{4}\cos 20^\circ \cos 80^\circ = \frac{1}{8} [2\cos 80^\circ \cos 20^\circ] = \frac{1}{8} [\cos 100^\circ + \cos 60^\circ] = \frac{1}{8}\cos 100^\circ + \frac{1}{16}$$
Notice that $$\cos 100^\circ = \cos(180^\circ - 80^\circ) = -\cos 80^\circ$$.
$$\text{LHS} = \frac{1}{8}\cos 80^\circ + \frac{1}{8}(-\cos 80^\circ) + \frac{1}{16} = 0 + \frac{1}{16} = \frac{1}{16}$$