Question

Evaluate the sum: \[ \tan^2 \frac{\pi}{16} + \tan^2 \frac{2\pi}{16} + \tan^2 \frac{3\pi}{16} + \tan^2 \frac{4\pi}{16} + \tan^2 \frac{5\pi}{16} + \tan^2 \frac{6\pi}{16} + \tan^2 \frac{7\pi}{16} \]

• A. \( 35 \)
• B. \( 41 \)
• C. \( 37 \)
• D. \( 33 \)

Hint:

For trigonometric summations, use angle sum identities and symmetry properties to simplify.

Answer 1

The Correct Option is A

To evaluate the sum: \[ \tan^2 \frac{\pi}{16} + \tan^2 \frac{2\pi}{16} + \tan^2 \frac{3\pi}{16} + \tan^2 \frac{4\pi}{16} + \tan^2 \frac{5\pi}{16} + \tan^2 \frac{6\pi}{16} + \tan^2 \frac{7\pi}{16}, \] we proceed as follows: Step 1: Simplify the angles The angles are symmetric around \( \frac{\pi}{2} \). Specifically: 
- \( \tan^2 \frac{\pi}{16} = \tan^2 \frac{15\pi}{16} \), 
- \( \tan^2 \frac{2\pi}{16} = \tan^2 \frac{14\pi}{16} \), 
- \( \tan^2 \frac{3\pi}{16} = \tan^2 \frac{13\pi}{16} \), 
- \( \tan^2 \frac{4\pi}{16} = \tan^2 \frac{\pi}{4} = 1 \), 
- \( \tan^2 \frac{5\pi}{16} = \tan^2 \frac{11\pi}{16} \), 
- \( \tan^2 \frac{6\pi}{16} = \tan^2 \frac{10\pi}{16} \), 
- \( \tan^2 \frac{7\pi}{16} = \tan^2 \frac{9\pi}{16} \). Thus, the sum can be rewritten as: \[ 2 \left( \tan^2 \frac{\pi}{16} + \tan^2 \frac{2\pi}{16} + \tan^2 \frac{3\pi}{16} \right) + 1. \] Step 2: Use the identity for \( \tan^2 \theta \) The identity for \( \tan^2 \theta \) is: \[ \tan^2 \theta = \frac{1 - \cos 2\theta}{1 + \cos 2\theta}. \] However, this approach is not straightforward for this problem. Instead, we use a known result for the sum of squares of tangent functions: \[ \sum_{k=1}^{n-1} \tan^2 \frac{k\pi}{2n} = \frac{(2n-1)(2n-2)}{6}. \] For \( n = 8 \), the sum becomes: \[ \sum_{k=1}^{7} \tan^2 \frac{k\pi}{16} = \frac{(16-1)(16-2)}{6} = \frac{15 \cdot 14}{6} = 35. \] Step 3: Verify the result The sum of the squares of the tangent functions is: \[ \tan^2 \frac{\pi}{16} + \tan^2 \frac{2\pi}{16} + \tan^2 \frac{3\pi}{16} + \tan^2 \frac{4\pi}{16} + \tan^2 \frac{5\pi}{16} + \tan^2 \frac{6\pi}{16} + \tan^2 \frac{7\pi}{16} = 35. \] Final Answer: \[ \boxed{35} \]