Evaluate `sin^(2)26^(@)+sin^(2)64^(@)`.

Answer

To evaluate \( \sin^2(26^\circ) + \sin^2(64^\circ) \), we can follow these steps:

### Step 1: Rewrite \( \sin^2(64^\circ) \)
We know that \( \sin(64^\circ) \) can be expressed in terms of cosine:
\[ \sin(64^\circ) = \cos(90^\circ - 64^\circ) = \cos(26^\circ) \]
Thus, we can rewrite \( \sin^2(64^\circ) \) as:
\[ \sin^2(64^\circ) = \cos^2(26^\circ) \]

### Step 2: Substitute into the original expression
Now we can substitute this back into the original expression:
\[ \sin^2(26^\circ) + \sin^2(64^\circ) = \sin^2(26^\circ) + \cos^2(26^\circ) \]

### Step 3: Apply the Pythagorean identity
We can use the Pythagorean identity, which states that:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
Applying this identity to our expression:
\[ \sin^2(26^\circ) + \cos^2(26^\circ) = 1 \]

### Final Answer
Thus, the value of \( \sin^2(26^\circ) + \sin^2(64^\circ) \) is:
\[ \boxed{1} \]

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