Question

*(p). $\sin \frac{\pi}{13} + \sin \frac{3\pi}{13} + \sin \frac{4\pi}{13} = \frac{1}{2} \cdot \sqrt{\frac{13 + 3\sqrt{13}}{2}}$.

Answer 1

True

Answer 2

The identity is indeed true.

Let $S = \sin \frac{\pi}{13} + \sin \frac{3\pi}{13} + \sin \frac{4\pi}{13}$. This sum is a specific value related to the properties of the 13th roots of unity, particularly the Gaussian periods. For a prime $p \equiv 1 \pmod 4$, the sum of sines of certain angles can be expressed in terms of $\sqrt{p}$. For $p=13$, it is a known result that:

$\sin \frac{\pi}{13} + \sin \frac{3\pi}{13} + \sin \frac{4\pi}{13} = \frac{1}{2} \sqrt{\frac{13+3\sqrt{13}}{2}}$

And also:

$\sin \frac{2\pi}{13} + \sin \frac{5\pi}{13} + \sin \frac{6\pi}{13} = \frac{1}{2} \sqrt{\frac{13-3\sqrt{13}}{2}}$

These results are derived using advanced number theory and properties of cyclotomic polynomials and Gaussian sums. The direct derivation is beyond the scope of typical JEE/NEET syllabus but the identity itself is a known mathematical fact. Therefore, the statement is true.