Find the value of
`sin^(2)1^(@)+sin^(2)2^(@)+sin^(2)3^(@)+ ----+sin^(2)87^(@)+sin^(2)88^(@)+sin^(2)89^(@)+sin^(2)90^(@)`.

Options

A. `44(1)/(2)`
B. 44
C. `45(1)/(2)`
D. N.O.T.

Answer

To find the value of the expression

\[ \sin^2(1^\circ) + \sin^2(2^\circ) + \sin^2(3^\circ) + \ldots + \sin^2(89^\circ) + \sin^2(90^\circ) \]

we can use the property of sine function and symmetry.

### Step 1: Use the identity for sine
We know that:

\[ \sin(90^\circ - x) = \cos(x) \]

This means that:

\[ \sin^2(90^\circ - x) = \cos^2(x) \]

### Step 2: Pair the terms
We can pair the terms in the sum:

\[ \sin^2(1^\circ) + \sin^2(89^\circ), \quad \sin^2(2^\circ) + \sin^2(88^\circ), \quad \ldots, \quad \sin^2(44^\circ) + \sin^2(46^\circ) \]

And we have the middle term:

\[ \sin^2(45^\circ) = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2} \]

### Step 3: Calculate the pairs
Each pair can be simplified using the identity:

\[ \sin^2(x) + \sin^2(90^\circ - x) = \sin^2(x) + \cos^2(x) = 1 \]

Thus, we have:

\[ \sin^2(1^\circ) + \sin^2(89^\circ) = 1 \]
\[ \sin^2(2^\circ) + \sin^2(88^\circ) = 1 \]
\[ \sin^2(3^\circ) + \sin^2(87^\circ) = 1 \]
\[ \ldots \]
\[ \sin^2(44^\circ) + \sin^2(46^\circ) = 1 \]

### Step 4: Count the pairs
We have 44 pairs (from \(1^\circ\) to \(44^\circ\)), and we also have the middle term \(\sin^2(45^\circ)\).

So, the total sum is:

\[ 44 \cdot 1 + \frac{1}{2} = 44 + 0.5 = 44.5 \]

### Final Answer
Thus, the value of

\[ \sin^2(1^\circ) + \sin^2(2^\circ) + \ldots + \sin^2(90^\circ) = 44.5 \]

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