let `f(x)=e^(cos^(-1)sin(x + pi/3))` then `f((8pi)/9) and f((-7pi)/4)`

Options

A. `f((8pi)/9)=e^((5pi)/18)`
B. `f((8pi)/9)=e^((13pi)/18)`
C. `f(-(7pi)/4)=e^((pi)/12)`
D. `f(-(7pi)/9)=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 the values \( x = \frac{8\pi}{9} \) and \( x = -\frac{7\pi}{4} \).

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

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

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. Now, find \( \sin\left(\frac{11\pi}{9}\right) \):
\[ \sin\left(\frac{11\pi}{9}\right) = \sin\left(\pi + \frac{2\pi}{9}\right) = -\sin\left(\frac{2\pi}{9}\right) \]

4. 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 \( \cos^{-1}(-x) = \pi - \cos^{-1}(x) \):
\[ f\left(\frac{8\pi}{9}\right) = e^{\pi - \cos^{-1}(\sin\left(\frac{2\pi}{9}\right))} \]

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

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: Calculate \( f\left(-\frac{7\pi}{4}\right) \)

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

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

3. Now, find \( \sin\left(-\frac{17\pi}{12}\right) \):
\[ \sin\left(-\frac{17\pi}{12}\right) = -\sin\left(\frac{17\pi}{12}\right) \]

4. 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 \( \cos^{-1}(-x) = \pi - \cos^{-1}(x) \):
\[ f\left(-\frac{7\pi}{4}\right) = e^{\pi - \cos^{-1}(\sin\left(\frac{17\pi}{12}\right))} \]

6. Since \( \sin\left(\frac{17\pi}{12}\right) = \cos\left(\frac{\pi}{2} - \frac{17\pi}{12}\right) \):
\[ \cos^{-1}(\sin\left(\frac{17\pi}{12}\right)) = \frac{\pi}{2} - \frac{17\pi}{12} = \frac{6\pi}{12} - \frac{17\pi}{12} = -\frac{11\pi}{12} \]

7. Thus, we have:
\[ f\left(-\frac{7\pi}{4}\right) = e^{\pi - \left(-\frac{11\pi}{12}\right)} = e^{\pi + \frac{11\pi}{12}} = e^{\frac{12\pi}{12} + \frac{11\pi}{12}} = e^{\frac{23\pi}{12}} \]

### Final Results:
- \( f\left(\frac{8\pi}{9}\right) = e^{\frac{13\pi}{18}} \)
- \( f\left(-\frac{7\pi}{4}\right) = e^{\frac{23\pi}{12}} \)