Question ID: #747
If $\frac{\cos^{2}48^{\circ}-\sin^{2}12^{\circ}}{\sin^{2}24^{\circ}-\sin^{2}6^{\circ}}=\frac{\alpha+\beta\sqrt{5}}{2}$ where $\alpha, \beta \in \mathbb{N}$, then $\alpha+\beta$ is equal to
Solution:
Using the identities:
1. $\cos^2 A – \sin^2 B = \cos(A+B)\cos(A-B)$
2. $\sin^2 A – \sin^2 B = \sin(A+B)\sin(A-B)$
Numerator:
$$ \cos^2 48^\circ – \sin^2 12^\circ = \cos(48^\circ+12^\circ)\cos(48^\circ-12^\circ) = \cos 60^\circ \cos 36^\circ $$
Denominator:
$$ \sin^2 24^\circ – \sin^2 6^\circ = \sin(24^\circ+6^\circ)\sin(24^\circ-6^\circ) = \sin 30^\circ \sin 18^\circ $$
The expression becomes:
$$ E = \frac{\cos 60^\circ \cos 36^\circ}{\sin 30^\circ \sin 18^\circ} $$
Substitute the standard values:
$\cos 60^\circ = \frac{1}{2}, \quad \sin 30^\circ = \frac{1}{2}$
$\cos 36^\circ = \frac{\sqrt{5}+1}{4}, \quad \sin 18^\circ = \frac{\sqrt{5}-1}{4}$
$$ E = \frac{\frac{1}{2} \cdot \frac{\sqrt{5}+1}{4}}{\frac{1}{2} \cdot \frac{\sqrt{5}-1}{4}} $$
$$ E = \frac{\sqrt{5}+1}{\sqrt{5}-1} $$
Rationalize the denominator:
$$ E = \frac{(\sqrt{5}+1)^2}{(\sqrt{5}-1)(\sqrt{5}+1)} = \frac{5 + 1 + 2\sqrt{5}}{5-1} $$
$$ E = \frac{6 + 2\sqrt{5}}{4} = \frac{3 + \sqrt{5}}{2} $$
Comparing with $\frac{\alpha + \beta\sqrt{5}}{2}$:
$\alpha = 3, \quad \beta = 1$.
We need $\alpha + \beta$:
$$ \alpha + \beta = 3 + 1 = 4 $$
Ans. (4)
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