If ^2 48^ - ^2 12^^2 24^ - ^2 6^ = + √52, where , N, then + is equal to ___________ Answer…

Question

If $\dfrac{\cos^2 48^\circ - \sin^2 12^\circ}{\sin^2 24^\circ - \sin^2 6^\circ} = \dfrac{\alpha + \beta\sqrt5}{2}$, where $\alpha,\beta \in \mathbb{N}$, then $\alpha+\beta$ is equal to \_\_\_\_\_\_\_\_\_\_\_

Answer: 4

Accepted Answer

Using $\cos^2A - \sin^2B = \cos(A+B)\cos(A-B)$ and $\sin^2A - \sin^2B = \sin(A+B)\sin(A-B)$:

$$\cos^2 48^\circ - \sin^2 12^\circ = \cos 60^\circ \cos 36^\circ, \qquad \sin^2 24^\circ - \sin^2 6^\circ = \sin 30^\circ \sin 18^\circ$$

With $\cos 60^\circ = \tfrac12$, $\cos 36^\circ = \tfrac{\sqrt5+1}{4}$, $\sin 30^\circ = \tfrac12$, $\sin 18^\circ = \tfrac{\sqrt5-1}{4}$:

$$\frac{\cos 60^\circ\cos 36^\circ}{\sin 30^\circ \sin 18^\circ} = \frac{\sqrt5+1}{\sqrt5-1} = \frac{(\sqrt5+1)^2}{4} = \frac{6+2\sqrt5}{4} = \frac{3+\sqrt5}{2}$$

So $\alpha = 3,\ \beta = 1$, giving $\alpha+\beta = 4$.

Solutions · 1