If `a+b = 3- cos 4 theta` and `a-b =4 sin 2theta`, then ab is always less than or equal to

Answer

To solve the problem, we start with the equations given:

1. \( a + b = 3 - \cos(4\theta) \) (Equation 1)
2. \( a - b = 4 \sin(2\theta) \) (Equation 2)

We need to find the product \( ab \) and determine the maximum value it can take.

### Step 1: Add the two equations

Adding Equation 1 and Equation 2:

\[ (a + b) + (a - b) = (3 - \cos(4\theta)) + (4 \sin(2\theta) \]

This simplifies to:

\[ 2a = 3 - \cos(4\theta) + 4 \sin(2\theta) \]

### Step 2: Solve for \( a \)

Now, we can solve for \( a \):

\[ a = \frac{3 - \cos(4\theta) + 4 \sin(2\theta)}{2} \]

### Step 3: Subtract the two equations

Next, we subtract Equation 2 from Equation 1:

\[ (a + b) - (a - b) = (3 - \cos(4\theta)) - (4 \sin(2\theta)) \]

This simplifies to:

\[ 2b = 3 - \cos(4\theta) - 4 \sin(2\theta) \]

### Step 4: Solve for \( b \)

Now, we can solve for \( b \):

\[ b = \frac{3 - \cos(4\theta) - 4 \sin(2\theta)}{2} \]

### Step 5: Find the product \( ab \)

Now we can find the product \( ab \):

\[ ab = \left(\frac{3 - \cos(4\theta) + 4 \sin(2\theta)}{2}\right) \left(\frac{3 - \cos(4\theta) - 4 \sin(2\theta)}{2}\right) \]

This can be simplified using the identity \( (x + y)(x - y) = x^2 - y^2 \):

Let \( x = 3 - \cos(4\theta) \) and \( y = 4 \sin(2\theta) \):

\[ ab = \frac{(3 - \cos(4\theta))^2 - (4 \sin(2\theta))^2}{4} \]

### Step 6: Expand and simplify

Expanding this gives:

\[ ab = \frac{(3 - \cos(4\theta))^2 - 16 \sin^2(2\theta)}{4} \]

### Step 7: Use the identity for \( \cos(4\theta) \)

Using the identity \( \cos(4\theta) = 1 - 2\sin^2(2\theta) \):

Substituting this into the equation:

\[ ab = \frac{(3 - (1 - 2\sin^2(2\theta)))^2 - 16 \sin^2(2\theta)}{4} \]

This simplifies to:

\[ ab = \frac{(2 + 2\sin^2(2\theta))^2 - 16 \sin^2(2\theta)}{4} \]

### Step 8: Further simplification

Let \( z = \sin^2(2\theta) \):

\[ ab = \frac{(2 + 2z)^2 - 16z}{4} = \frac{4(1 + z)^2 - 16z}{4} = (1 + z)^2 - 4z \]

### Step 9: Final expression

This can be further simplified to:

\[ ab = 1 + 2z + z^2 - 4z = 1 - 2z + z^2 \]

### Step 10: Analyze the quadratic

The expression \( z^2 - 2z + 1 = (z - 1)^2 \) is always non-negative and achieves a maximum value of 0 when \( z = 1 \).

### Conclusion

Thus, the maximum value of \( ab \) occurs when \( z = 1 \) (i.e., \( \sin^2(2\theta) = 1 \)), leading to:

\[ ab \leq 1 \]

### Final Answer

Therefore, \( ab \) is always less than or equal to 1.

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