(a) Without using a calculator, determine positive integers $$ m $$ and $$ n $$ for which $$ \sin ^{6} 1^{\circ}+\sin ^{6} 2^{\circ}+\sin ^{6} 3^{\circ}+\cdots+\sin ^{6} 87^{\circ}+\sin ^{6} 88^{\circ}+\sin ^{6} 89^{\circ}=\frac{m}{n} $$ (The sum on the left side of the equation consists of 89 terms of the form $$ \sin ^{6} x^{\circ} $$, where $$ x $$ takes each positive integer value from 1 to 89 .) (b) Let $$ f(n) $$ be the number of positive integers that have exactly $$ n $$ digits and whose digits have a sum of 5 . Determine, with proof, how many of the 2014 integers $$ f(1), f(2), \ldots, f(2014) $$ have a units digit of 1 .

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Problem (a) asks us to evaluate the sum:

$$
S = \sin^6 1^{\circ} + \sin^6 2^{\circ} + \cdots + \sin^6 89^{\circ}
$$

and express it in the form $$\frac{m}{n}$$ where $$m$$ and $$n$$ are positive integers.

### Step 1: Express $$\sin^6 x$$ Using Multiple Angle Identities

To simplify $$S$$, we start by expressing $$\sin^6 x$$ in terms of multiple angles. Using trigonometric identities, we have:

$$
\sin^6 x = \frac{1}{32} \left(10 - 15 \cos 2x + 6 \cos 4x - \cos 6x\right)
$$

### Step 2: Substitute and Simplify the Sum $$S$$

Substituting the expression for $$\sin^6 x$$ into the sum $$S$$, we get:

$$
S = \sum_{k=1}^{89} \sin^6 k^{\circ} = \frac{1}{32} \left( \sum_{k=1}^{89} 10 - 15 \sum_{k=1}^{89} \cos 2k^{\circ} + 6 \sum_{k=1}^{89} \cos 4k^{\circ} - \sum_{k=1}^{89} \cos 6k^{\circ} \right)
$$

### Step 3: Evaluate Each Sum Individually

1. **Constant Term:**
   
   $$
   \sum_{k=1}^{89} 10 = 10 \times 89 = 890
   $$

2. **Sum of $$\cos 2k^{\circ}$$:**
   
   Using the formula for the sum of cosine terms:
   
   $$
   \sum_{k=1}^{89} \cos 2k^{\circ} = 0
   $$
   
   (This is because the cosine terms cancel out over the interval.)

3. **Sum of $$\cos 4k^{\circ}$$:**
   
   $$
   \sum_{k=1}^{89} \cos 4k^{\circ} = -1
   $$
   
   (This result arises from evaluating the sum using multiple-angle identities.)

4. **Sum of $$\cos 6k^{\circ}$$:**
   
   $$
   \sum_{k=1}^{89} \cos 6k^{\circ} = 0
   $$
   
   (Similar to the sum of $$\cos 2k^{\circ}$$, the terms cancel out.)

### Step 4: Combine the Results

Substituting these results back into the expression for $$S$$:

$$
S = \frac{1}{32} \left(890 - 15 \times 0 + 6 \times (-1) - 0\right) = \frac{1}{32} (890 - 6) = \frac{884}{32}
$$

Simplifying the fraction:

$$
\frac{884}{32} = \frac{221}{8}
$$

### Conclusion

Therefore, the sum $$S$$ can be expressed as:

$$
S = \frac{221}{8}
$$

So, the positive integers $$m$$ and $$n$$ are $$m = 221$$ and $$n = 8$$.

**Answer:**  
Problem a Answer:

The sum is equal to 221⁄8. So, m = 221 and n = 8.
The sum $$ S = \sin^6 1^{\circ} + \sin^6 2^{\circ} + \cdots + \sin^6 89^{\circ} $$ equals $$ \frac{221}{8} $$. Therefore, $$ m = 221 $$ and $$ n = 8 $$.

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