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:

Use trigonometric identities and properties to simplify and solve the expression.

Answer 1

The Correct Option is C

To solve this problem, we need to simplify the given expression:

\[\frac{\cos^2 48^\circ - \sin^2 12^\circ}{\sin^2 24^\circ - \sin^2 6^\circ}\]

and equate it to 

\[\frac{\alpha + \sqrt{5\beta}}{2}\]

, then find the value of  

\[(\alpha + \beta)\]

.

Let's break it down step-by-step:

1. Use the identity: 

\[\cos^2 A - \sin^2 B = \cos(A + B) \cos(A - B)\]

1. .
2. Apply the identity to the numerator:
   • We have: 

\[\cos^2 48^\circ - \sin^2 12^\circ = \cos(48^\circ + 12^\circ) \cos(48^\circ - 12^\circ)\]

• .
• Calculate: 

\[\cos(48^\circ + 12^\circ) = \cos 60^\circ = \frac{1}{2}\]

•  and 

\[\cos(48^\circ - 12^\circ) = \cos 36^\circ\]

• .

1. Thus, the numerator becomes: 

\[\frac{1}{2} \cos 36^\circ\]

1. .
2. For the denominator, use the identity: 

\[\sin^2 A - \sin^2 B = \sin(A+B) \sin(A-B)\]

1. .
2. Apply this to the denominator:
   • We have: 

\[\sin^2 24^\circ - \sin^2 6^\circ = \sin(24^\circ + 6^\circ) \sin(24^\circ - 6^\circ)\]

• .
• Calculate: 

\[\sin(24^\circ + 6^\circ) = \sin 30^\circ = \frac{1}{2}\]

•  and 

\[\sin(24^\circ - 6^\circ) = \sin 18^\circ\]

• .

1. So, the denominator becomes: 

\[\frac{1}{2} \sin 18^\circ\]

1. .
2. Now substitute back into the main fraction:
   • We get: 

\[\frac{\frac{1}{2} \cos 36^\circ}{\frac{1}{2} \sin 18^\circ}\]

• .
• Simplify it to: 

\[\frac{\cos 36^\circ}{\sin 18^\circ}\]

• .

1. Use the identities for angles related to 18 and 36 degrees:
   • We know: 

\[\cos 36^\circ = \sin 54^\circ\]

•  and 

\[\sin 18^\circ = \frac{\sqrt{5}-1}{4}\]

• .
• This gives: 

\[\frac{\sin 54^\circ}{\frac{\sqrt{5}-1}{4}}\]

• .

1. Further simplify:
   • We have: 

\[\sin 54^\circ = \cos 36^\circ = \frac{\sqrt{5} + 1}{4}\]

• .
• Therefore, 

\[\frac{\frac{\sqrt{5} + 1}{4}}{\frac{\sqrt{5} - 1}{4}} = \frac{\sqrt{5} + 1}{\sqrt{5} - 1}\]

• .
• Multiply the numerator and denominator by the conjugate to rationalize:
• This becomes: 

\[\frac{\sqrt{5} + 1}{\sqrt{5} - 1} \times \frac{\sqrt{5} + 1}{\sqrt{5} + 1} = \frac{(\sqrt{5} + 1)^2}{(\sqrt{5})^2 - (1)^2}\]

• .
• Simplify: \(\frac{(\sqrt{5} + 1)^2}{5 - 1} = \frac{6 + 2\sqrt{5}}{4} = \frac{3}{2} + \frac{\sqrt{5}}{2}\)

1. Compare with the given expression: 

\[\frac{\alpha + \sqrt{5\beta}}{2}\]

1. :
   • We equate: 

\[\alpha = 3\]

•  and 

\[\beta = 1\]

• .
• Thus, \((\alpha + \beta) = 3 + 1 = 4\).

Therefore, the value of \((\alpha + \beta)\) is 4.