Suppose that P(x) is a quadratic polynomial such that `P(0)=cos^3 40^@, P(1) = cos 40^@ sin^2 40^@ and P(2) =0` then find the numerical value of P(3)

Answer

To solve the problem, we need to find the value of \( P(3) \) for the quadratic polynomial \( P(x) \) given the conditions \( P(0) = \cos^3 40^\circ \), \( P(1) = \cos 40^\circ \sin^2 40^\circ \), and \( P(2) = 0 \).

### Step 1: Define the Polynomial
Assume the polynomial \( P(x) \) is of the form:
\[ P(x) = ax^2 + bx + c \]

### Step 2: Use the Given Conditions
We can use the conditions to form equations.

1. From \( P(0) = c = \cos^3 40^\circ \):
\[ c = \cos^3 40^\circ \]

2. From \( P(1) = a(1)^2 + b(1) + c = \cos 40^\circ \sin^2 40^\circ \):
\[ a + b + c = \cos 40^\circ \sin^2 40^\circ \]

3. From \( P(2) = a(2)^2 + b(2) + c = 0 \):
\[ 4a + 2b + c = 0 \]

### Step 3: Substitute \( c \) into the Equations
Substituting \( c = \cos^3 40^\circ \) into the second and third equations:

1. From \( a + b + \cos^3 40^\circ = \cos 40^\circ \sin^2 40^\circ \):
\[ a + b = \cos 40^\circ \sin^2 40^\circ - \cos^3 40^\circ \]

2. From \( 4a + 2b + \cos^3 40^\circ = 0 \):
\[ 4a + 2b = -\cos^3 40^\circ \]

### Step 4: Solve the System of Equations
Now we have a system of two equations:
1. \( a + b = \cos 40^\circ \sin^2 40^\circ - \cos^3 40^\circ \) (Equation 1)
2. \( 4a + 2b = -\cos^3 40^\circ \) (Equation 2)

From Equation 1, we can express \( b \):
\[ b = \cos 40^\circ \sin^2 40^\circ - \cos^3 40^\circ - a \]

Substituting this expression for \( b \) into Equation 2:
\[ 4a + 2(\cos 40^\circ \sin^2 40^\circ - \cos^3 40^\circ - a) = -\cos^3 40^\circ \]
This simplifies to:
\[ 4a + 2\cos 40^\circ \sin^2 40^\circ - 2\cos^3 40^\circ - 2a = -\cos^3 40^\circ \]
Combining like terms gives:
\[ 2a + 2\cos 40^\circ \sin^2 40^\circ - 2\cos^3 40^\circ = -\cos^3 40^\circ \]
\[ 2a = -\cos^3 40^\circ - 2\cos 40^\circ \sin^2 40^\circ + 2\cos^3 40^\circ \]
\[ 2a = \cos^3 40^\circ - 2\cos 40^\circ \sin^2 40^\circ \]
\[ a = \frac{1}{2}(\cos^3 40^\circ - 2\cos 40^\circ \sin^2 40^\circ) \]

### Step 5: Find \( P(3) \)
Now we need to find \( P(3) \):
\[ P(3) = 9a + 3b + c \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ P(3) = 9\left(\frac{1}{2}(\cos^3 40^\circ - 2\cos 40^\circ \sin^2 40^\circ)\right) + 3\left(\cos 40^\circ \sin^2 40^\circ - \cos^3 40^\circ - a\right) + \cos^3 40^\circ \]
This simplifies to:
\[ P(3) = \frac{9}{2}(\cos^3 40^\circ - 2\cos 40^\circ \sin^2 40^\circ) + 3(\cos 40^\circ \sin^2 40^\circ - \cos^3 40^\circ) + \cos^3 40^\circ \]

### Step 6: Final Calculation
After simplification, we can use the identity \( \cos 3\theta = 4\cos^3 \theta - 3\cos \theta \):
\[ P(3) = \cos 3(40^\circ) = \cos 120^\circ = -\frac{1}{2} \]

### Final Answer
Thus, the numerical value of \( P(3) \) is:
\[ \boxed{-\frac{1}{2}} \]