If `a cos^ 3 3theta + b cos^4 theta = 16 cos^6 theta + 9 cos^2 theta` is an identity then-

Options

A. `a=1, b=24`
B. `a=3, b=24`
C. `a=4, b=2`
D. `a=7, b=18`

Answer

To solve the equation \( a \cos^3(3\theta) + b \cos^4(\theta) = 16 \cos^6(\theta) + 9 \cos^2(\theta) \) and find the values of \( a \) and \( b \), we will follow these steps:

### Step 1: Set up the equation
We start with the equation given in the problem:
\[ a \cos^3(3\theta) + b \cos^4(\theta) = 16 \cos^6(\theta) + 9 \cos^2(\theta) \]

### Step 2: Substitute \( \theta = 0 \)
Substituting \( \theta = 0 \):
- \( \cos(0) = 1 \)
- \( \cos(3 \cdot 0) = 1 \)

The equation becomes:
\[ a \cdot 1^3 + b \cdot 1^4 = 16 \cdot 1^6 + 9 \cdot 1^2 \]
This simplifies to:
\[ a + b = 16 + 9 \]
Thus, we have:
\[ a + b = 25 \quad \text{(Equation 1)} \]

### Step 3: Substitute \( \theta = \frac{\pi}{3} \)
Next, we substitute \( \theta = \frac{\pi}{3} \):
- \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \)
- \( \cos(3 \cdot \frac{\pi}{3}) = \cos(\pi) = -1 \)

The equation becomes:
\[ a \cdot (-1)^3 + b \cdot \left(\frac{1}{2}\right)^4 = 16 \cdot \left(\frac{1}{2}\right)^6 + 9 \cdot \left(\frac{1}{2}\right)^2 \]
This simplifies to:
\[ -a + b \cdot \frac{1}{16} = 16 \cdot \frac{1}{64} + 9 \cdot \frac{1}{4} \]
Calculating the right side:
\[ 16 \cdot \frac{1}{64} = \frac{1}{4} \quad \text{and} \quad 9 \cdot \frac{1}{4} = \frac{9}{4} \]
So, we have:
\[ -a + \frac{b}{16} = \frac{1}{4} + \frac{9}{4} = \frac{10}{4} = \frac{5}{2} \]
Multiplying through by 16 to eliminate the fraction:
\[ -16a + b = 40 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations
Now we have two equations:
1. \( a + b = 25 \)
2. \( -16a + b = 40 \)

We can solve these equations simultaneously. From Equation 1, we express \( b \):
\[ b = 25 - a \]
Substituting this into Equation 2:
\[ -16a + (25 - a) = 40 \]
This simplifies to:
\[ -17a + 25 = 40 \]
Rearranging gives:
\[ -17a = 40 - 25 \]
\[ -17a = 15 \]
\[ a = -\frac{15}{17} \]

Now substituting \( a \) back into Equation 1 to find \( b \):
\[ -\frac{15}{17} + b = 25 \]
\[ b = 25 + \frac{15}{17} = \frac{425}{17} + \frac{15}{17} = \frac{440}{17} \]

### Final Values
Thus, the values of \( a \) and \( b \) are:
\[ a = -\frac{15}{17}, \quad b = \frac{440}{17} \]