Question

$\frac{\cos 15^\circ \cos^2 22\frac{1}{2}^\circ - \sin 75^\circ \sin^2 52\frac{1}{2}^\circ}{\cos^2 15^\circ - \cos^2 75^\circ} =$

• A. 1
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{1}{8}$

Hint:

When you see angles like $15^\circ, 75^\circ, 22.5^\circ, 67.5^\circ$, etc., look for opportunities to use co-function identities to relate sines and cosines, and half-angle or double-angle formulas. Simplifying expressions often involves turning sums into products or vice versa, and identities like $\cos^2 A - \sin^2 B$ are very useful.

Answer 1

The Correct Option is C

Step 1: Simplify Numerator and Denominator: Let \( N \) be the numerator and \( D \) be the denominator. Recall \( \sin 75^\circ = \cos 15^\circ \). \[ N = \cos 15^\circ \cos^2 22.5^\circ - \cos 15^\circ \sin^2 52.5^\circ \] \[ N = \cos 15^\circ (\cos^2 22.5^\circ - \sin^2 52.5^\circ) \] Using the identity \( \cos^2 A - \sin^2 B = \cos(A+B)\cos(A-B) \): Here \( A = 22.5^\circ \) and \( B = 52.5^\circ \). \[ A+B = 75^\circ, \quad A-B = -30^\circ \] \[ N = \cos 15^\circ [\cos(75^\circ)\cos(-30^\circ)] \] \[ N = \cos 15^\circ \sin 15^\circ \left(\frac{\sqrt{3}}{2}\right) \] Using \( 2\sin\theta\cos\theta = \sin 2\theta \): \[ N = \frac{1}{2} \sin(30^\circ) \frac{\sqrt{3}}{2} = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8} \]
Step 2: Simplify Denominator: \[ D = \cos^2 15^\circ - \sin^2 15^\circ \quad (\because \cos 75^\circ = \sin 15^\circ) \] \[ D = \cos(2 \times 15^\circ) = \cos 30^\circ = \frac{\sqrt{3}}{2} \]
Step 3: Calculate Ratio: \[ \frac{N}{D} = \frac{\sqrt{3}/8}{\sqrt{3}/2} = \frac{1}{4} \]