If the product `(sin1^@)(sin3^@)(sin5^@)(sin7^@)...(sin89^@)=(1/2^n)` then find the value of n

Answer

To solve the problem, we need to find the value of \( n \) such that:

\[ (\sin 1^\circ)(\sin 3^\circ)(\sin 5^\circ)(\sin 7^\circ) \ldots (\sin 89^\circ) = \frac{1}{2^n} \]

### Step-by-Step Solution:

1. **Identify the Product**:
The product consists of the sine of all odd degrees from \( 1^\circ \) to \( 89^\circ \).

\[ P = \sin 1^\circ \sin 3^\circ \sin 5^\circ \ldots \sin 89^\circ \]

2. **Use the Identity**:
We can use the identity \( \sin x = \sin(90^\circ - x) \). This means that:
\[ \sin 89^\circ = \sin 1^\circ, \quad \sin 87^\circ = \sin 3^\circ, \quad \ldots, \quad \sin 1^\circ = \sin 89^\circ \]
Thus, we can pair the terms:
\[ P = (\sin 1^\circ \sin 89^\circ)(\sin 3^\circ \sin 87^\circ) \ldots (\sin 45^\circ) \]

3. **Count the Terms**:
There are \( 45 \) terms in total (from \( 1^\circ \) to \( 89^\circ \) in odd increments).

4. **Simplify the Product**:
Each pair contributes \( \sin x \sin(90^\circ - x) = \sin x \cos x = \frac{1}{2} \sin(2x) \). Therefore, we can express the product as:
\[ P = \frac{1}{2^{44}} \cdot \sin 45^\circ \]
since there are \( 44 \) pairs and \( \sin 45^\circ = \frac{\sqrt{2}}{2} \).

5. **Final Expression**:
Thus, we have:
\[ P = \frac{1}{2^{44}} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2^{45}} \]

6. **Express in Terms of \( \frac{1}{2^n} \)**:
We want to express \( P \) as \( \frac{1}{2^n} \):
\[ \frac{\sqrt{2}}{2^{45}} = \frac{1}{2^{45 - \frac{1}{2}}} = \frac{1}{2^{44.5}} \]

7. **Conclusion**:
Therefore, we find that:
\[ n = 44.5 \]

### Final Answer:
\[ n = \frac{89}{2} \]