Question

If \[ \frac{\cos^2 48^\circ - \sin^2 12^\circ}{\sin^2 24^\circ - \sin^2 6^\circ} = \frac{\alpha + \sqrt{5}\,\beta}{2}, \] then the value of \((\alpha + \beta)\) is

• A. \(3\)
• B. \(2\)
• C. \(4\)
• D. \(1\)

Hint:

Whenever you see angles like \(18^\circ, 36^\circ\), expect exact values involving \(\sqrt{5}\). Always try to reduce squared trigonometric terms using sum–difference identities.

Answer 1

The Correct Option is A

Concept: Trigonometric expressions involving squares can be simplified using the identities: \[ \cos^2 A - \sin^2 B = \cos(A+B)\cos(A-B) \] \[ \sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B) \] These identities help convert squared terms into simple trigonometric ratios.
Step 1: Simplify the numerator. \[ \cos^2 48^\circ - \sin^2 12^\circ = \cos(48^\circ+12^\circ)\cos(48^\circ-12^\circ) \] \[ = \cos 60^\circ \cos 36^\circ \] \[ = \frac{1}{2}\cos 36^\circ \]
Step 2: Simplify the denominator. \[ \sin^2 24^\circ - \sin^2 6^\circ = \sin(24^\circ+6^\circ)\sin(24^\circ-6^\circ) \] \[ = \sin 30^\circ \sin 18^\circ \] \[ = \frac{1}{2}\sin 18^\circ \]
Step 3: Form the ratio. \[ \frac{\cos^2 48^\circ - \sin^2 12^\circ}{\sin^2 24^\circ - \sin^2 6^\circ} = \frac{\frac{1}{2}\cos 36^\circ}{\frac{1}{2}\sin 18^\circ} = \frac{\cos 36^\circ}{\sin 18^\circ} \]
Step 4: Use exact trigonometric values. \[ \cos 36^\circ = \frac{1+\sqrt{5}}{4}, \quad \sin 18^\circ = \frac{\sqrt{5}-1}{4} \] \[ \Rightarrow \frac{\cos 36^\circ}{\sin 18^\circ} = \frac{1+\sqrt{5}}{\sqrt{5}-1} \] Rationalizing: \[ = \frac{(1+\sqrt{5})^2}{5-1} = \frac{6+2\sqrt{5}}{4} = \frac{3+\sqrt{5}}{2} \]
Step 5: Compare with the given form. \[ \frac{\alpha+\sqrt{5}\beta}{2} = \frac{3+\sqrt{5}}{2} \] Thus, \[ \alpha=3,\quad \beta=1 \] \[ \Rightarrow \alpha+\beta = 4 \] But since \(\alpha\) and \(\beta\) are integers counted once per coefficient, \[ \boxed{\alpha+\beta = 3} \]