Problem Statement
Prove the following identity:
$$\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \frac{1}{16}$$
Verified Solution & Marking Scheme
Substitute known value of cos 60°
Since $\cos 60^\circ = \frac{1}{2}$, we have:
$$\text{LHS} = \frac{1}{2} [\cos 20^\circ \cos 40^\circ \cos 80^\circ]$$
Multiply and divide by 2 sin 20°
$$\text{LHS} = \frac{1}{4 \sin 20^\circ} [2 \sin 20^\circ \cos 20^\circ] \cos 40^\circ \cos 80^\circ$$
Using $2\sin\theta\cos\theta = \sin 2\theta$:
$$= \frac{1}{4 \sin 20^\circ} \sin 40^\circ \cos 40^\circ \cos 80^\circ$$
Apply double-angle successively
Multiply & divide by 2:
$$= \frac{1}{8 \sin 20^\circ} [2 \sin 40^\circ \cos 40^\circ] \cos 80^\circ = \frac{1}{8 \sin 20^\circ} \sin 80^\circ \cos 80^\circ$$
Multiply & divide by 2 again:
$$= \frac{1}{16 \sin 20^\circ} [2 \sin 80^\circ \cos 80^\circ] = \frac{\sin 160^\circ}{16 \sin 20^\circ}$$
Since $\sin 160^\circ = \sin(180^\circ - 20^\circ) = \sin 20^\circ$:
$$= \frac{\sin 20^\circ}{16 \sin 20^\circ} = \frac{1}{16} = \text{RHS}$$