If `f(x)=cos^-1(sin(4cos^2x-1)), ` then `1/pif'(pi/3).f(pi/10)` is

Answer

To solve the problem, we need to find the value of \( \frac{1}{\pi} f'(\frac{\pi}{3}) \cdot f(\frac{\pi}{10}) \) where \( f(x) = \cos^{-1}(\sin(4\cos^2 x - 1)) \).

### Step 1: Rewrite the function
We start with:
\[ f(x) = \cos^{-1}(\sin(4\cos^2 x - 1)) \]
We know that \( \sin(4\cos^2 x - 1) \) can be rewritten using the identity \( \sin(a) = \sin(\pi - a) \) to bring it into a more manageable form.

### Step 2: Simplify the argument of sine
We can express \( 4\cos^2 x - 1 \) in terms of sine:
\[ 4\cos^2 x - 1 = 3 - 4\sin^2 x \]
Thus, we can rewrite:
\[ f(x) = \cos^{-1}(\sin(3 - 4\sin^2 x)) \]

### Step 3: Use trigonometric identities
Using the identity \( \sin(3x) = 3\sin x - 4\sin^3 x \), we can express \( f(x) \) as:
\[ f(x) = \cos^{-1}(\sin(3x)) \]

### Step 4: Further simplification
We can express \( f(x) \) using the identity for inverse trigonometric functions:
\[ f(x) = \frac{\pi}{2} - \sin^{-1}(\sin(3x)) \]
Since \( \sin(3x) \) can take values outside the principal range, we adjust it:
\[ f(x) = \frac{\pi}{2} - (3x) \quad \text{(for } 3x \text{ in the range of } [0, \pi]) \]

### Step 5: Find \( f'(\frac{\pi}{3}) \)
Now we differentiate \( f(x) \):
\[ f'(x) = -3 \]
Thus:
\[ f'(\frac{\pi}{3}) = -3 \]

### Step 6: Calculate \( f(\frac{\pi}{10}) \)
Next, we calculate \( f(\frac{\pi}{10}) \):
\[ f(\frac{\pi}{10}) = \frac{\pi}{2} - 3 \cdot \frac{\pi}{10} = \frac{\pi}{2} - \frac{3\pi}{10} = \frac{5\pi}{10} - \frac{3\pi}{10} = \frac{2\pi}{10} = \frac{\pi}{5} \]

### Step 7: Combine results
Now we can compute:
\[ \frac{1}{\pi} f'(\frac{\pi}{3}) \cdot f(\frac{\pi}{10}) = \frac{1}{\pi} (-3) \cdot \frac{\pi}{5} = -\frac{3}{5} \]

### Final Answer
Thus, the final result is:
\[ \frac{1}{\pi} f'(\frac{\pi}{3}) \cdot f(\frac{\pi}{10}) = -\frac{3}{5} \]