If sinA/3 = 1/2(a + 1/a), prove that: sinA = -1/2(a^3 + 1/a^3)

We are given: $\sin \frac{A}{3} = \frac{1}{2}\left(a + \frac{1}{a}\right)$

We need to prove: $\sin A = -\frac{1}{2}\left(a^3 + \frac{1}{a^3}\right)$

Using the triple angle formula for sine: $\sin 3\theta = 3\sin \theta - 4\sin^3 \theta$

Let $\theta = \frac{A}{3}$, then: $\sin A = 3\sin \frac{A}{3} - 4\sin^3 \frac{A}{3}$

Substituting the given value: $\sin A = 3 \cdot \frac{1}{2}\left(a + \frac{1}{a}\right) - 4\left(\frac{1}{2}\left(a + \frac{1}{a}\right)\right)^3$

Simplifying: $\sin A = \frac{3}{2}\left(a + \frac{1}{a}\right) - \frac{4}{8}\left(a + \frac{1}{a}\right)^3$

$\sin A = \frac{3}{2}\left(a + \frac{1}{a}\right) - \frac{1}{2}\left(a + \frac{1}{a}\right)^3$

Expanding $\left(a + \frac{1}{a}\right)^3$: $\left(a + \frac{1}{a}\right)^3 = a^3 + 3a + \frac{3}{a} + \frac{1}{a^3}$

Substituting back: $\sin A = \frac{3}{2}\left(a + \frac{1}{a}\right) - \frac{1}{2}\left(a^3 + 3a + \frac{3}{a} + \frac{1}{a^3}\right)$

$\sin A = \frac{3}{2}a + \frac{3}{2a} - \frac{1}{2}a^3 - \frac{3}{2}a - \frac{3}{2a} - \frac{1}{2a^3}$

$\sin A = -\frac{1}{2}a^3 - \frac{1}{2a^3}$

$\sin A = -\frac{1}{2}\left(a^3 + \frac{1}{a^3}\right)$

Therefore, we have proved that: $\sin A = -\frac{1}{2}\left(a^3 + \frac{1}{a^3}\right)$