`(sin^(2)1^(@) + sin^(2)3^(@) + sin^(2)5^(@) +sin^(2)7^(@) + ….. + sin^(2)87^(@) + sin^(2)89^(@))` equals

Options

A. a)23
B. b)22
C. c)`22(1/2)`
D. d)`23(1/2)`

Answer

To solve the expression \( \sin^2 1^\circ + \sin^2 3^\circ + \sin^2 5^\circ + \sin^2 7^\circ + \ldots + \sin^2 89^\circ \), we can use the properties of complementary angles and some trigonometric identities.

### Step-by-Step Solution:

1. **Identify the Pattern**: The terms in the series are sine squares of odd degrees from 1 to 89. We can pair the terms based on their complementary angles. For example, \( \sin^2 1^\circ \) pairs with \( \sin^2 89^\circ \), \( \sin^2 3^\circ \) pairs with \( \sin^2 87^\circ \), and so on.

2. **Use the Complementary Angle Identity**: Recall that \( \sin^2 \theta + \cos^2 \theta = 1 \). Since \( \sin^2 89^\circ = \cos^2 1^\circ \), we can write:
\[ \sin^2 1^\circ + \sin^2 89^\circ = \sin^2 1^\circ + \cos^2 1^\circ = 1 \]
Similarly, we can pair:
\[ \sin^2 3^\circ + \sin^2 87^\circ = 1 \]
\[ \sin^2 5^\circ + \sin^2 85^\circ = 1 \]
\[ \ldots \]
\[ \sin^2 43^\circ + \sin^2 47^\circ = 1 \]
The last unpaired term is \( \sin^2 45^\circ \).

3. **Count the Pairs**: The odd degrees from 1 to 89 form an arithmetic progression where:
- First term \( a = 1 \)
- Last term \( l = 89 \)
- Common difference \( d = 2 \)

The number of terms \( n \) can be calculated using the formula:
\[ n = \frac{l - a}{d} + 1 = \frac{89 - 1}{2} + 1 = 45 \]
Since we can pair them, we have \( 22 \) pairs (each summing to \( 1 \)) and one unpaired term \( \sin^2 45^\circ \).

4. **Calculate the Total**: Each of the \( 22 \) pairs contributes \( 1 \) to the sum:
\[ 22 \times 1 + \sin^2 45^\circ \]
We know that \( \sin^2 45^\circ = \frac{1}{2} \). Thus, the total sum becomes:
\[ 22 + \frac{1}{2} = 22.5 \]

5. **Final Answer**: Therefore, the value of the expression is:
\[ \boxed{22.5} \]