Question

For \( A = \frac{\pi}{24} \), if \( \frac{\sin 2A + \sin 3A + \sin 4A}{\cos 2A + \cos 3A + \cos 4A} = k \), then \( (k+1)^2 = \)

• A. \( 0 \)
• B. \( 1 \)
• C. \( 2 \)
• D. \( 3 \)

Hint:

To simplify expressions like \( \frac{\sin nx + \sin (n+1)x + \sin (n+2)x}{\cos nx + \cos (n+1)x + \cos (n+2)x} \), the result is always \( \tan((n+1)x) \). This "middle angle" property is a common shortcut for symmetric sums.

Answer 1

The Correct Option is C

Concept: We use sum-to-product transformation formulas to simplify the trigonometric fraction:
• \( \sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) \)
• \( \cos C + \cos D = 2 \cos\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) \)

Step 1: Simplifying the expression for \( k \).
Group the first and third terms in both numerator and denominator: \[ k = \frac{(\sin 4A + \sin 2A) + \sin 3A}{(\cos 4A + \cos 2A) + \cos 3A} = \frac{2 \sin 3A \cos A + \sin 3A}{2 \cos 3A \cos A + \cos 3A} \] Factor out \( \sin 3A \) and \( \cos 3A \): \[ k = \frac{\sin 3A (2 \cos A + 1)}{\cos 3A (2 \cos A + 1)} = \tan 3A \]

Step 2: Evaluating \( k \) for \( A = \pi/24 \).
\[ 3A = 3 \times \frac{\pi}{24} = \frac{\pi}{8} \] \[ k = \tan\left(\frac{\pi}{8}\right) = \sqrt{2} - 1 \]

Step 3: Calculating \( (k+1)^2 \).
\[ k + 1 = (\sqrt{2} - 1) + 1 = \sqrt{2} \] \[ (k+1)^2 = (\sqrt{2})^2 = 2 \]