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 need to simplify a trigonometric expression and express it in the form \(\frac{\alpha + \beta\sqrt{5}}{\gamma}\). Then we have to find the sum of the integers \(\alpha, \beta, \gamma\).

Step 2: Key Formula or Approach:
We will use the following trigonometric identities:
\(\cos^2 A - \sin^2 B = \cos(A+B)\cos(A-B)\)
\(\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)\) We will also 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 \(\cos^2 A - \sin^2 B = \cos(A+B)\cos(A-B)\): \[ \cos(48^\circ+12^\circ)\cos(48^\circ-12^\circ) = \cos(60^\circ)\cos(36^\circ) \] We know \(\cos 60^\circ = \frac{1}{2}\) and \(\cos 36^\circ = \frac{\sqrt{5}+1}{4}\). \[ \text{Numerator} = \frac{1}{2} \cdot \frac{\sqrt{5}+1}{4} = \frac{\sqrt{5}+1}{8} \] Denominator: \(\sin^2 24^\circ - \sin^2 6^\circ\) Using the identity \(\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)\): \[ \sin(24^\circ+6^\circ)\sin(24^\circ-6^\circ) = \sin(30^\circ)\sin(18^\circ) \] We know \(\sin 30^\circ = \frac{1}{2}\) and \(\sin 18^\circ = \frac{\sqrt{5}-1}{4}\). \[ \text{Denominator} = \frac{1}{2} \cdot \frac{\sqrt{5}-1}{4} = \frac{\sqrt{5}-1}{8} \] 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 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: \(\alpha = 3\), \(\beta = 1\), \(\gamma = 2\).
These are natural numbers and in the lowest form.
The required value is \(\alpha + \beta + \gamma = 3 + 1 + 2 = 6\).