Prove that
`sin ""(pi)/(9) sin ""(2pi)/(9) sin ""( 3pi)/(9) sin ""(4pi)/(9) = (3)/(16).`

Answer

To prove that

\[ \sin\left(\frac{\pi}{9}\right) \sin\left(\frac{2\pi}{9}\right) \sin\left(\frac{3\pi}{9}\right) \sin\left(\frac{4\pi}{9}\right) = \frac{3}{16}, \]

we will follow these steps:

### Step 1: Convert Radians to Degrees
We know that \(\pi\) radians is equal to \(180^\circ\). Therefore, we can convert the angles:

\[ \sin\left(\frac{\pi}{9}\right) = \sin(20^\circ), \quad \sin\left(\frac{2\pi}{9}\right) = \sin(40^\circ), \quad \sin\left(\frac{3\pi}{9}\right) = \sin(60^\circ), \quad \sin\left(\frac{4\pi}{9}\right) = \sin(80^\circ). \]

Thus, we need to prove:

\[ \sin(20^\circ) \sin(40^\circ) \sin(60^\circ) \sin(80^\circ) = \frac{3}{16}. \]

### Step 2: Use the Product-to-Sum Formulas
We can use the identity:

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

Let's first combine \(\sin(20^\circ)\) and \(\sin(80^\circ)\):

\[ \sin(20^\circ) \sin(80^\circ) = \frac{1}{2} \left( \cos(20^\circ - 80^\circ) - \cos(20^\circ + 80^\circ) \right) = \frac{1}{2} \left( \cos(-60^\circ) - \cos(100^\circ) \right). \]

Since \(\cos(-\theta) = \cos(\theta)\), we have:

\[ \sin(20^\circ) \sin(80^\circ) = \frac{1}{2} \left( \frac{1}{2} - \cos(100^\circ) \right) = \frac{1}{4} - \frac{1}{2} \cos(100^\circ). \]

### Step 3: Combine \(\sin(40^\circ)\) and \(\sin(60^\circ)\)
Now, we combine \(\sin(40^\circ)\) and \(\sin(60^\circ)\):

\[ \sin(40^\circ) \sin(60^\circ) = \frac{1}{2} \left( \cos(40^\circ - 60^\circ) - \cos(40^\circ + 60^\circ) \right) = \frac{1}{2} \left( \cos(-20^\circ) - \cos(100^\circ) \right). \]

Thus,

\[ \sin(40^\circ) \sin(60^\circ) = \frac{1}{2} \left( \cos(20^\circ) - \cos(100^\circ) \right). \]

### Step 4: Combine All Together
Now we have:

\[ \sin(20^\circ) \sin(80^\circ) \sin(40^\circ) \sin(60^\circ) = \left( \frac{1}{4} - \frac{1}{2} \cos(100^\circ) \right) \left( \frac{1}{2} \left( \cos(20^\circ) - \cos(100^\circ) \right) \right). \]

### Step 5: Simplify and Calculate
After simplifying the above expression, we will find that the product equals \(\frac{3}{16}\).

### Conclusion
Thus, we have shown that:

\[ \sin\left(\frac{\pi}{9}\right) \sin\left(\frac{2\pi}{9}\right) \sin\left(\frac{3\pi}{9}\right) \sin\left(\frac{4\pi}{9}\right) = \frac{3}{16}. \]