`alphacos^2 3theta+betacos^4theta=16cos^6theta+9cos^2theta`. Find `alpha & beta` if its a identity

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 \( \alpha \cos^2 3\theta + \beta \cos^4 \theta = 16 \cos^6 \theta + 9 \cos^2 \theta \) and find the values of \( \alpha \) and \( \beta \) such that it holds as an identity, we can follow these steps:

### Step 1: Substitute Values for \(\theta\)

1. **Let \(\theta = 0\)**:
- Calculate \( \cos 0 = 1 \).
- Substitute into the equation:
\[ \alpha \cos^2(0) + \beta \cos^4(0) = 16 \cos^6(0) + 9 \cos^2(0) \]
- This simplifies to:
\[ \alpha(1) + \beta(1) = 16(1) + 9(1) \]
\[ \alpha + \beta = 25 \quad \text{(Equation 1)} \]

### Step 2: Substitute Another Value for \(\theta\)

2. **Let \(\theta = \pi\)**:
- Calculate \( \cos \pi = -1 \).
- Substitute into the equation:
\[ \alpha \cos^2(3\pi) + \beta \cos^4(\pi) = 16 \cos^6(\pi) + 9 \cos^2(\pi) \]
- This simplifies to:
\[ \alpha(1) + \beta(1) = 16(1) + 9(1) \]
\[ \alpha + \beta = 25 \quad \text{(Equation 2)} \]

### Step 3: Substitute a Third Value for \(\theta\)

3. **Let \(\theta = \frac{\pi}{3}\)**:
- Calculate \( \cos \frac{\pi}{3} = \frac{1}{2} \).
- Substitute into the equation:
\[ \alpha \cos^2(3 \cdot \frac{\pi}{3}) + \beta \cos^4(\frac{\pi}{3}) = 16 \cos^6(\frac{\pi}{3}) + 9 \cos^2(\frac{\pi}{3}) \]
- This simplifies to:
\[ \alpha \cos^2(\pi) + \beta \left(\frac{1}{2}\right)^4 = 16 \left(\frac{1}{2}\right)^6 + 9 \left(\frac{1}{2}\right)^2 \]
- Which becomes:
\[ \alpha(1) + \beta \left(\frac{1}{16}\right) = 16 \left(\frac{1}{64}\right) + 9 \left(\frac{1}{4}\right) \]
- Simplifying further:
\[ \alpha + \frac{\beta}{16} = \frac{16}{64} + \frac{9}{4} \]
\[ \alpha + \frac{\beta}{16} = \frac{1}{4} + \frac{9}{4} = \frac{10}{4} = \frac{5}{2} \quad \text{(Equation 3)} \]

### Step 4: Solve the Equations

Now we have the following equations:
1. \( \alpha + \beta = 25 \) (Equation 1)
2. \( \alpha + \frac{\beta}{16} = \frac{5}{2} \) (Equation 3)

From Equation 1, we can express \( \alpha \) in terms of \( \beta \):
\[ \alpha = 25 - \beta \]

Substituting into Equation 3:
\[ 25 - \beta + \frac{\beta}{16} = \frac{5}{2} \]
Multiplying through by 16 to eliminate the fraction:
\[ 16(25 - \beta) + \beta = 40 \]
\[ 400 - 16\beta + \beta = 40 \]
\[ 400 - 15\beta = 40 \]
\[ 15\beta = 400 - 40 \]
\[ 15\beta = 360 \]
\[ \beta = 24 \]

### Step 5: Find \(\alpha\)

Substituting \( \beta = 24 \) back into Equation 1:
\[ \alpha + 24 = 25 \]
\[ \alpha = 1 \]

### Final Answer

Thus, the values of \( \alpha \) and \( \beta \) are:
\[ \alpha = 1, \quad \beta = 24 \]