Find the value of `sin^(2)5^(@)+ sin^(2)10^(@)+ sin^(2)15^(@)+* * * +sin^(2)90^(@)`.

Options

A. 8
B. 9
C. `(17)/(2)`
D. `(19)/(2)`

Explanation

To find the value of \( \sin^2 5^\circ + \sin^2 10^\circ + \sin^2 15^\circ + \ldots + \sin^2 90^\circ \), we can use the properties of sine and some known identities.

### Step 1: Identify the Series
The series consists of \( \sin^2 \) values from \( 5^\circ \) to \( 90^\circ \) in increments of \( 5^\circ \). The angles are:
\[ 5^\circ, 10^\circ, 15^\circ, \ldots, 90^\circ \]

### Step 2: Count the Terms
The angles form an arithmetic series where:
- First term \( a = 5^\circ \)
- Last term \( l = 90^\circ \)
- Common difference \( d = 5^\circ \)

To find the number of terms \( n \):
\[ n = \frac{l - a}{d} + 1 = \frac{90 - 5}{5} + 1 = \frac{85}{5} + 1 = 17 + 1 = 18 \]

### Step 3: Use the Identity for Sine
We can use the identity:
\[ \sin^2 x = \frac{1 - \cos(2x)}{2} \]
Thus, we can rewrite the series:
\[ \sin^2 5^\circ + \sin^2 10^\circ + \sin^2 15^\circ + \ldots + \sin^2 90^\circ = \sum_{k=1}^{18} \sin^2(5k^\circ) \]

### Step 4: Apply the Identity
Using the identity, we have:
\[ \sum_{k=1}^{18} \sin^2(5k^\circ) = \sum_{k=1}^{18} \frac{1 - \cos(10k^\circ)}{2} \]
This can be simplified to:
\[ \frac{1}{2} \sum_{k=1}^{18} (1 - \cos(10k^\circ)) = \frac{1}{2} \left( \sum_{k=1}^{18} 1 - \sum_{k=1}^{18} \cos(10k^\circ) \right) \]

### Step 5: Calculate the First Part
The first part is straightforward:
\[ \sum_{k=1}^{18} 1 = 18 \]

### Step 6: Calculate the Cosine Sum
Now we need to calculate \( \sum_{k=1}^{18} \cos(10k^\circ) \). This is a finite geometric series:
\[ \sum_{k=1}^{n} \cos(a + (k-1)d) = \frac{\sin\left(\frac{nd}{2}\right) \cos\left(a + \frac{(n-1)d}{2}\right)}{\sin\left(\frac{d}{2}\right)} \]
Here, \( n = 18 \), \( a = 10^\circ \), and \( d = 10^\circ \):
\[ \sum_{k=1}^{18} \cos(10k^\circ) = \frac{\sin\left(90^\circ\right) \cos\left(10^\circ + 85^\circ\right)}{\sin(5^\circ)} = \frac{1 \cdot \cos(95^\circ)}{\sin(5^\circ)} = \frac{-\sin(5^\circ)}{\sin(5^\circ)} = -1 \]

### Step 7: Substitute Back
Now substituting back:
\[ \sum_{k=1}^{18} \sin^2(5k^\circ) = \frac{1}{2} \left( 18 - (-1) \right) = \frac{1}{2} \cdot 19 = 9.5 \]

### Final Answer
Thus, the value of \( \sin^2 5^\circ + \sin^2 10^\circ + \sin^2 15^\circ + \ldots + \sin^2 90^\circ \) is:
\[ \boxed{9.5} \]