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 Standard Angle
We know that $\cos 60^\circ = \frac{1}{2}$. Therefore:
$$\text{LHS} = \frac{1}{2} [\cos 20^\circ \cos 40^\circ \cos 80^\circ]$$
Apply Product-to-Sum Formula on Cos 40° and Cos 20°
Multiply and divide by 2:
$$\text{LHS} = \frac{1}{4} [2 \cos 40^\circ \cos 20^\circ] \cos 80^\circ$$
Using $2\cos A \cos B = \cos(A + B) + \cos(A - B)$:
$$2 \cos 40^\circ \cos 20^\circ = \cos 60^\circ + \cos 20^\circ = \frac{1}{2} + \cos 20^\circ$$
$$\text{LHS} = \frac{1}{4} \left( \frac{1}{2} + \cos 20^\circ \right) \cos 80^\circ = \frac{1}{8} \cos 80^\circ + \frac{1}{4} \cos 80^\circ \cos 20^\circ$$
Simplify and Cancel Complementary Terms
Again multiply and divide the second term by 2:
$$\frac{1}{4} \cos 80^\circ \cos 20^\circ = \frac{1}{8} [2 \cos 80^\circ \cos 20^\circ] = \frac{1}{8} [\cos 100^\circ + \cos 60^\circ]$$
$$\text{LHS} = \frac{1}{8} \cos 80^\circ + \frac{1}{8} \cos 100^\circ + \frac{1}{8} \left(\frac{1}{2}\right)$$
Since $\cos 100^\circ = \cos(180^\circ - 80^\circ) = -\cos 80^\circ$:
$$\frac{1}{8} \cos 80^\circ + \frac{1}{8}(-\cos 80^\circ) = 0$$
$$\text{LHS} = 0 + \frac{1}{16} = \frac{1}{16} = \text{RHS}$$.