Prove that in a ` A B C ,sin^2A+sin^2B+sin^2C<=9/4dot`

Explanation

To prove that in triangle \( ABC \), \( \sin^2 A + \sin^2 B + \sin^2 C \leq \frac{9}{4} \), we can follow these steps:

### Step 1: Use the identity for \( \sin^2 \theta \)
We know that:
\[ \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \]
Using this identity, we can express \( \sin^2 A \), \( \sin^2 B \), and \( \sin^2 C \) as follows:
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