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: Understanding the Concept
The problem involves simplifying a complex trigonometric expression. The key is to use co-function identities and other trigonometric formulas to simplify the numerator and denominator separately.
Step 2: Key Formula or Approach
We will use the following trigonometric identities: 1. Co-function identities: $\sin(90^\circ - x) = \cos x$ and $\cos(90^\circ - x) = \sin x$. 2. Double angle identity: $\cos^2 A - \sin^2 A = \cos(2A)$. 3. Product-to-sum/difference related identity: $\cos^2 A - \cos^2 B = \sin^2 B - \sin^2 A = \sin(A+B)\sin(B-A)$. 4. Double angle identity for sine: $2\sin A \cos A = \sin(2A)$.
Step 3: Detailed Explanation
Let's simplify the numerator and denominator separately. Numerator: $N = \cos 15^\circ \cos^2 22.5^\circ - \sin 75^\circ \sin^2 52.5^\circ$. First, use co-function identities: $\sin 75^\circ = \sin(90^\circ - 15^\circ) = \cos 15^\circ$. Also, $\sin 52.5^\circ = \sin(90^\circ - 37.5^\circ) = \cos 37.5^\circ$. Substitute these into the numerator: \[ N = \cos 15^\circ \cos^2 22.5^\circ - \cos 15^\circ \cos^2 37.5^\circ \] Factor out $\cos 15^\circ$: \[ N = \cos 15^\circ (\cos^2 22.5^\circ - \cos^2 37.5^\circ) \] Now use the identity $\cos^2 A - \cos^2 B = \sin(B+A)\sin(B-A)$: \[ N = \cos 15^\circ [\sin(37.5^\circ + 22.5^\circ)\sin(37.5^\circ - 22.5^\circ)] \] \[ N = \cos 15^\circ [\sin(60^\circ)\sin(15^\circ)] \] Rearrange the terms: \[ N = \sin 15^\circ \cos 15^\circ \sin 60^\circ \] Use the double angle identity for sine, $2\sin A\cos A = \sin(2A)$: \[ N = \frac{1}{2}(2\sin 15^\circ \cos 15^\circ) \sin 60^\circ = \frac{1}{2} \sin(2 \times 15^\circ) \sin 60^\circ \] \[ N = \frac{1}{2} \sin 30^\circ \sin 60^\circ = \frac{1}{2} \left(\frac{1}{2}\right) \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{8} \] Denominator: $D = \cos^2 15^\circ - \cos^2 75^\circ$. Use the co-function identity $\cos 75^\circ = \cos(90^\circ - 15^\circ) = \sin 15^\circ$. \[ D = \cos^2 15^\circ - \sin^2 15^\circ \] Use the double angle identity for cosine, $\cos^2 A - \sin^2 A = \cos(2A)$: \[ D = \cos(2 \times 15^\circ) = \cos 30^\circ = \frac{\sqrt{3}}{2} \] Final Expression: Now, divide the simplified numerator by the simplified denominator: \[ \frac{N}{D} = \frac{\sqrt{3}/8}{\sqrt{3}/2} = \frac{\sqrt{3}}{8} \times \frac{2}{\sqrt{3}} = \frac{2}{8} = \frac{1}{4} \] Step 4: Final Answer
The value of the expression is $\frac{1}{4}$.