Question

The value of $4 \cos 36^{\circ}\cos 72^{\circ}$ is equal to

• A. 1
• B. 2
• C. $\sqrt{2}$
• D. 2
• E. 0

Hint:

Logic Tip: A useful identity to memorize for competitive exams is $\cos \theta \cos(60^{\circ}-\theta) \cos(60^{\circ}+\theta) = \frac{1}{4} \cos 3\theta$. Alternatively, memorizing $\cos 36^{\circ}\cos 72^{\circ} = \frac{1}{4}$ will save you significant calculation time.

Answer 1

The Correct Option is A

Concept:
This problem relies on knowing the standard exact trigonometric values for $36^{\circ}$ and $72^{\circ}$ (or $18^{\circ}$). The key values are: $$\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$$ $$\cos 72^{\circ} = \sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$$
Step 1: Substitute the exact values into the expression.
We need to evaluate $4 \cos 36^{\circ} \cos 72^{\circ}$. Substitute the known values: $$= 4 \left( \frac{\sqrt{5}+1}{4} \right) \left( \frac{\sqrt{5}-1}{4} \right)$$
Step 2: Simplify the product of the fractions.
Factor out the denominators: $$= 4 \cdot \frac{(\sqrt{5}+1)(\sqrt{5}-1)}{16}$$
Step 3: Apply the difference of squares formula.
The numerator is in the form $(a+b)(a-b) = a^2 - b^2$, where $a = \sqrt{5}$ and $b = 1$: $$(\sqrt{5})^2 - (1)^2 = 5 - 1 = 4$$
Step 4: Calculate the final value.
Substitute the simplified numerator back into the expression: $$= 4 \cdot \frac{4}{16}$$ $$= \frac{16}{16} = 1$$