Tangent lines are drawn at the points P and Q where f''(x) vanishes for the function f(x)=cos x on `[0,2pi]` The tangent lines at P and Q intersect each other at R so as to form a triangle PQR. If area triangle PQR is `kpi^(2)` then find the value of 36k.

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Identify where \( f''(x) \) vanishes
Given the function \( f(x) = \cos x \), we need to find the second derivative and determine where it is equal to zero.

1. First, we find the first derivative:
\[ f'(x) = -\sin x \]

2. Next, we find the second derivative:
\[ f''(x) = -\cos x \]

3. Set the second derivative equal to zero to find the points where it vanishes:
\[ -\cos x = 0 \implies \cos x = 0 \]
The solutions in the interval \([0, 2\pi]\) are:
\[ x = \frac{\pi}{2}, \frac{3\pi}{2} \]

### Step 2: Find the points \( P \) and \( Q \)
The points \( P \) and \( Q \) correspond to \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \).

1. Calculate the coordinates of point \( P \):
\[ P = \left( \frac{\pi}{2}, f\left(\frac{\pi}{2}\right) \right) = \left( \frac{\pi}{2}, 0 \right) \]

2. Calculate the coordinates of point \( Q \):
\[ Q = \left( \frac{3\pi}{2}, f\left(\frac{3\pi}{2}\right) \right) = \left( \frac{3\pi}{2}, 0 \right) \]

### Step 3: Find the equations of the tangent lines at \( P \) and \( Q \)
1. Calculate the slopes at points \( P \) and \( Q \):
- At \( P \):
\[ f' \left( \frac{\pi}{2} \right) = -\sin\left(\frac{\pi}{2}\right) = -1 \]
- At \( Q \):
\[ f' \left( \frac{3\pi}{2} \right) = -\sin\left(\frac{3\pi}{2}\right) = 1 \]

2. Write the equation of the tangent line at \( P \):
\[ y - 0 = -1\left(x - \frac{\pi}{2}\right) \implies y = -x + \frac{\pi}{2} \]

3. Write the equation of the tangent line at \( Q \):
\[ y - 0 = 1\left(x - \frac{3\pi}{2}\right) \implies y = x - \frac{3\pi}{2} \]

### Step 4: Find the intersection point \( R \) of the tangent lines
Set the equations equal to find \( R \):
\[ -x + \frac{\pi}{2} = x - \frac{3\pi}{2} \]
Solving for \( x \):
\[ \frac{\pi}{2} + \frac{3\pi}{2} = 2x \implies 2\pi = 2x \implies x = \pi \]
Substituting \( x = \pi \) into one of the tangent equations to find \( y \):
\[ y = -\pi + \frac{\pi}{2} = -\frac{\pi}{2} \]
Thus, the coordinates of point \( R \) are:
\[ R = \left( \pi, -\frac{\pi}{2} \right) \]

### Step 5: Calculate the area of triangle \( PQR \)
The area \( A \) of triangle \( PQR \) can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
The base \( PQ \) is the distance between \( P \) and \( Q \):
\[ PQ = \frac{3\pi}{2} - \frac{\pi}{2} = \pi \]
The height is the vertical distance from point \( R \) to the line \( y = 0 \):
\[ \text{height} = \left| -\frac{\pi}{2} - 0 \right| = \frac{\pi}{2} \]
Thus, the area is:
\[ A = \frac{1}{2} \times \pi \times \frac{\pi}{2} = \frac{\pi^2}{4} \]

### Step 6: Relate the area to \( k \)
According to the problem, the area is given as \( k\pi^2 \):
\[ \frac{\pi^2}{4} = k\pi^2 \implies k = \frac{1}{4} \]

### Step 7: Find \( 36k \)
Finally, calculate \( 36k \):
\[ 36k = 36 \times \frac{1}{4} = 9 \]

### Final Answer
Thus, the value of \( 36k \) is \( \boxed{9} \).

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