Question

The value of $\cos(18^\circ - A) \cos(18^\circ + A) - \cos(72^\circ - A) \cos(72^\circ + A)$ is equal to

• A. $\cos 54^\circ$
• B. $\cos 36^\circ$
• C. $\sin 54^\circ$
• D. $\sin 36^\circ$

Answer 1

Answer: (B)

$\cos 36^\circ$

Answer 2

Using the product-to-sum formula:

$$\cos(18^\circ - A)\cos(18^\circ + A) = \cos^2 18^\circ - \sin^2 A$$

and similarly,

$$\cos(72^\circ - A)\cos(72^\circ + A) = \cos^2 72^\circ - \sin^2 A.$$

Subtracting the two:

$$[\cos^2 18^\circ - \sin^2 A] - [\cos^2 72^\circ - \sin^2 A] = \cos^2 18^\circ - \cos^2 72^\circ.$$

Noting that $\cos 72^\circ = \sin 18^\circ$, we have:

$$\cos^2 18^\circ - \cos^2 72^\circ = \cos^2 18^\circ - \sin^2 18^\circ = \cos 36^\circ,$$

where the last equality follows from the double angle formula $\cos 2\theta = \cos^2\theta - \sin^2\theta$ (with $\theta = 18^\circ$).