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 $\cos 100^\circ = \cos(180^\circ - 80^\circ) = -\cos 80^\circ$:
$\frac{1}{8} \cos 80^\circ + \frac{1}{8}(-\cos 80^\circ) = 0$
$\text{LHS} = 0 + \frac{1}{16} = \frac{1}{16} = \text{RHS}$.