Prove that: `"sin"20^0"sin"40^0"sin"60^0"sin"80^0=3/(16)`

Explanation

To prove that \( \sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ = \frac{3}{16} \), we can use some trigonometric identities and properties.

### Step-by-step Solution:

1. **Use the identity for sine of complementary angles**:
We know that \( \sin(90^\circ - x) = \cos x \). Therefore, we can rewrite \( \sin 80^\circ \) as:
\[
\sin 80^\circ = \cos 10^\circ
...