If `sinA=(336)/(625)where450^(@)ltAlt540^(@), then sin""(A)/(4)=`

Options

A. `3//5`
B. `-3//5`
C. `4//5`
D. `-4//5`

Answer

To solve the problem, we need to find the value of \(\sin\left(\frac{A}{4}\right)\) given that \(\sin A = \frac{336}{625}\) and \(450^\circ < A < 540^\circ\).

### Step-by-Step Solution:

1. **Identify the Quadrant for Angle A:**
Since \(A\) is in the range \(450^\circ < A < 540^\circ\), it lies in the second quadrant where sine is positive and cosine is negative.

2. **Calculate \(\cos A\):**
We use the identity:
\[ \cos^2 A = 1 - \sin^2 A \]
First, calculate \(\sin^2 A\):
\[ \sin^2 A = \left(\frac{336}{625}\right)^2 = \frac{112896}{390625} \]
Now, substitute this into the cosine formula:
\[ \cos^2 A = 1 - \frac{112896}{390625} = \frac{390625 - 112896}{390625} = \frac{277729}{390625} \]
Taking the square root, since cosine is negative in the second quadrant:
\[ \cos A = -\sqrt{\frac{277729}{390625}} = -\frac{527}{625} \]

3. **Use the Half-Angle Formula for Cosine:**
The half-angle formula states:
\[ \cos A = 2 \cos^2\left(\frac{A}{2}\right) - 1 \]
Setting this equal to our value of \(\cos A\):
\[ -\frac{527}{625} = 2 \cos^2\left(\frac{A}{2}\right) - 1 \]
Rearranging gives:
\[ 2 \cos^2\left(\frac{A}{2}\right) = -\frac{527}{625} + 1 = \frac{625 - 527}{625} = \frac{98}{625} \]
Thus,
\[ \cos^2\left(\frac{A}{2}\right) = \frac{49}{625} \]
Taking the square root:
\[ \cos\left(\frac{A}{2}\right) = \frac{7}{25} \]

4. **Determine the Sign of \(\cos\left(\frac{A}{2}\right)\):**
Since \(A\) is between \(450^\circ\) and \(540^\circ\), \(\frac{A}{2}\) will be between \(225^\circ\) and \(270^\circ\), which is in the third quadrant where cosine is negative:
\[ \cos\left(\frac{A}{2}\right) = -\frac{7}{25} \]

5. **Use the Half-Angle Formula for Sine:**
The sine half-angle formula is:
\[ \sin^2\left(\frac{A}{2}\right) = 1 - \cos^2\left(\frac{A}{2}\right) \]
Substituting our value:
\[ \sin^2\left(\frac{A}{2}\right) = 1 - \left(-\frac{7}{25}\right)^2 = 1 - \frac{49}{625} = \frac{576}{625} \]
Taking the square root:
\[ \sin\left(\frac{A}{2}\right) = \frac{24}{25} \]

6. **Calculate \(\sin\left(\frac{A}{4}\right)\):**
Using the half-angle formula again:
\[ \sin\left(\frac{A}{4}\right) = \sqrt{\frac{1 - \cos\left(\frac{A}{2}\right)}{2}} = \sqrt{\frac{1 - \left(-\frac{7}{25}\right)}{2}} = \sqrt{\frac{1 + \frac{7}{25}}{2}} = \sqrt{\frac{\frac{32}{25}}{2}} = \sqrt{\frac{32}{50}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

### Final Answer:
\[ \sin\left(\frac{A}{4}\right) = \frac{4}{5} \]