Let `f(x)=e^(cos^(-1)sin(x+ pi/3))`, then

Options

A. `f((8 pi)/(9))= e^((5pi)/(18))`
B. `f((8pi)/(9))= e^((13pi)/(18))`
C. `f(-(7pi)/(4))=e^((pi)/(12))`
D. `f(-(7pi)/(4))=e^((11pi)/(12))`

Answer

To solve the problem, we need to evaluate the function \( f(x) = e^{\cos^{-1}(\sin(x + \frac{\pi}{3}))} \) for \( x = \frac{8\pi}{9} \) and \( x = -\frac{7\pi}{4} \).

### Step 1: Evaluate \( f\left(\frac{8\pi}{9}\right) \)

1. **Substitute \( x = \frac{8\pi}{9} \) into the function:**
\[ f\left(\frac{8\pi}{9}\right) = e^{\cos^{-1}(\sin(\frac{8\pi}{9} + \frac{\pi}{3}))} \]

2. **Calculate \( \frac{8\pi}{9} + \frac{\pi}{3} \):**
\[ \frac{\pi}{3} = \frac{3\pi}{9} \quad \Rightarrow \quad \frac{8\pi}{9} + \frac{3\pi}{9} = \frac{11\pi}{9} \]

3. **Find \( \sin\left(\frac{11\pi}{9}\right) \):**
\[ \sin\left(\frac{11\pi}{9}\right) = -\sin\left(\frac{11\pi}{9} - \pi\right) = -\sin\left(\frac{2\pi}{9}\right) \]

4. **Now substitute back into the function:**
\[ f\left(\frac{8\pi}{9}\right) = e^{\cos^{-1}(-\sin\left(\frac{2\pi}{9}\right))} \]

5. **Use the identity \( \cos^{-1}(-y) = \pi - \cos^{-1}(y) \):**
\[ f\left(\frac{8\pi}{9}\right) = e^{\pi - \cos^{-1}(\sin\left(\frac{2\pi}{9}\right))} \]

6. **Since \( \cos^{-1}(\sin\theta) = \frac{\pi}{2} - \theta \):**
\[ \cos^{-1}(\sin\left(\frac{2\pi}{9}\right)) = \frac{\pi}{2} - \frac{2\pi}{9} = \frac{9\pi}{18} - \frac{4\pi}{18} = \frac{5\pi}{18} \]

7. **Thus, we have:**
\[ f\left(\frac{8\pi}{9}\right) = e^{\pi - \frac{5\pi}{18}} = e^{\frac{18\pi}{18} - \frac{5\pi}{18}} = e^{\frac{13\pi}{18}} \]

### Step 2: Evaluate \( f\left(-\frac{7\pi}{4}\right) \)

1. **Substitute \( x = -\frac{7\pi}{4} \) into the function:**
\[ f\left(-\frac{7\pi}{4}\right) = e^{\cos^{-1}(\sin(-\frac{7\pi}{4} + \frac{\pi}{3}))} \]

2. **Calculate \( -\frac{7\pi}{4} + \frac{\pi}{3} \):**
\[ -\frac{7\pi}{4} + \frac{\pi}{3} = -\frac{21\pi}{12} + \frac{4\pi}{12} = -\frac{17\pi}{12} \]

3. **Find \( \sin\left(-\frac{17\pi}{12}\right) \):**
\[ \sin\left(-\frac{17\pi}{12}\right) = -\sin\left(\frac{17\pi}{12}\right) \]

4. **Now substitute back into the function:**
\[ f\left(-\frac{7\pi}{4}\right) = e^{\cos^{-1}(-\sin\left(\frac{17\pi}{12}\right))} \]

5. **Use the identity \( \cos^{-1}(-y) = \pi - \cos^{-1}(y) \):**
\[ f\left(-\frac{7\pi}{4}\right) = e^{\pi - \cos^{-1}(\sin\left(\frac{17\pi}{12}\right))} \]

6. **Now find \( \cos^{-1}(\sin\left(\frac{17\pi}{12}\right)) \):**
- Since \( \frac{17\pi}{12} = \pi + \frac{5\pi}{12} \), we have:
\[ \sin\left(\frac{17\pi}{12}\right) = -\sin\left(\frac{5\pi}{12}\right) \]
- Thus:
\[ \cos^{-1}(-\sin\left(\frac{5\pi}{12}\right)) = \pi - \left(\frac{\pi}{2} - \frac{5\pi}{12}\right) = \frac{5\pi}{12} \]

7. **Thus, we have:**
\[ f\left(-\frac{7\pi}{4}\right) = e^{\pi - \frac{5\pi}{12}} = e^{\frac{12\pi}{12} - \frac{5\pi}{12}} = e^{\frac{7\pi}{12}} \]

### Final Results

- \( f\left(\frac{8\pi}{9}\right) = e^{\frac{13\pi}{18}} \)
- \( f\left(-\frac{7\pi}{4}\right) = e^{\frac{7\pi}{12}} \)