Question

If the value of \(\frac{\cos^2 48^\circ - \sin^2 12^\circ}{\sin^2 24^\circ - \sin^2 6^\circ}\) is \(\frac{\alpha + \beta\sqrt{5}}{\gamma}\) then value of \((\alpha + \beta + \gamma)\) (where \(\alpha, \beta, \gamma \in \mathbb{N}\) and are in lowest form) :

• A. 3
• B. 4
• C. 5
• D. 6

Hint:

Memorizing product-to-sum formulas and specific values like \(\sin 18^\circ\) and \(\cos 36^\circ\) is crucial for solving such trigonometry problems quickly in competitive exams.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are tasked with simplifying a trigonometric expression and expressing it in the form \( \frac{\alpha + \beta\sqrt{5}}{\gamma} \). After simplifying, we need to find the sum of the integers \( \alpha \), \( \beta \), and \( \gamma \).

Step 2: Key Formula or Approach:
We will use the following trigonometric identities:
\( \cos^2 A - \sin^2 B = \cos(A+B)\cos(A-B) \)
\( \sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B) \)
Additionally, we need the values of some standard angles:
\( \cos 36^\circ = \frac{\sqrt{5}+1}{4} \) and \( \sin 18^\circ = \frac{\sqrt{5}-1}{4} \).

Step 3: Detailed Explanation:
Let's simplify the numerator and the denominator separately.

Numerator: \( \cos^2 48^\circ - \sin^2 12^\circ \)
Using the identity \( \cos^2 A - \sin^2 B = \cos(A+B)\cos(A-B) \), we get: \[ \cos(48^\circ + 12^\circ) \cos(48^\circ - 12^\circ) = \cos(60^\circ) \cos(36^\circ) \] We know that \( \cos 60^\circ = \frac{1}{2} \) and \( \cos 36^\circ = \frac{\sqrt{5}+1}{4} \), so the numerator becomes: \[ \text{Numerator} = \frac{1}{2} \cdot \frac{\sqrt{5}+1}{4} = \frac{\sqrt{5}+1}{8} \]

Denominator: \( \sin^2 24^\circ - \sin^2 6^\circ \)
Using the identity \( \sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B) \), we get: \[ \sin(24^\circ + 6^\circ) \sin(24^\circ - 6^\circ) = \sin(30^\circ) \sin(18^\circ) \] We know that \( \sin 30^\circ = \frac{1}{2} \) and \( \sin 18^\circ = \frac{\sqrt{5}-1}{4} \), so the denominator becomes: \[ \text{Denominator} = \frac{1}{2} \cdot \frac{\sqrt{5}-1}{4} = \frac{\sqrt{5}-1}{8} \]

Now, let's find the value of the given expression: \[ \frac{\text{Numerator}}{\text{Denominator}} = \frac{(\sqrt{5}+1)/8}{(\sqrt{5}-1)/8} = \frac{\sqrt{5}+1}{\sqrt{5}-1} \] To rationalize the denominator, we multiply both the numerator and denominator by \( (\sqrt{5}+1) \): \[ \frac{\sqrt{5}+1}{\sqrt{5}-1} \times \frac{\sqrt{5}+1}{\sqrt{5}+1} = \frac{(\sqrt{5}+1)^2}{(\sqrt{5})^2 - 1^2} = \frac{5 + 1 + 2\sqrt{5}}{5-1} = \frac{6 + 2\sqrt{5}}{4} \] Simplifying the expression by dividing by 2: \[ \frac{3 + \sqrt{5}}{2} \]

Step 4: Final Answer:
We are given that the value is \( \frac{\alpha + \beta\sqrt{5}}{\gamma} \). Comparing this with our result \( \frac{3 + 1\sqrt{5}}{2} \), we get: \[ \alpha = 3, \quad \beta = 1, \quad \gamma = 2. \] These are natural numbers and in the lowest form. The required value is: \[ \alpha + \beta + \gamma = 3 + 1 + 2 = 6. \]