Let’s solve the given equation step by step:

The equation is: $\sqrt{9 - 8 \sin 50^\circ} = a + b \cos 50^\circ$ where $a$ and $b$ are integers. We need to find the product $ab$.

Step 1: Simplify the left-hand side

• First, evaluate $9 - 8 \sin 50^\circ$.
• Approximating $\sin 50^\circ \approx 0.766$, we get: $9 - 8 \sin 50^\circ \approx 9 - 8(0.766) = 9 - 6.128 = 2.872.$

Step 2: Calculate the square root

$\sqrt{2.872} \approx 1.694.$ Thus, the left-hand side becomes approximately $1.694$.

Step 3: Evaluate $\cos 50^\circ$

Approximating $\cos 50^\circ \approx 0.643$, the equation becomes: $1.694 \approx a + b(0.643).$

Step 4: Find integer values of $a$ and $b$

To satisfy the equation, try different integer values for $a$ and $b$ such that: $1.694 \approx a + 0.643b.$ Testing values:

• Let $a = 1$ and $b = 1$: $1 + 0.643 \approx 1.643 \quad (\text{close to } 1.694).$ So, $a = 1$ and $b = 1$ seem to work.

Step 5: Calculate the product $ab$

$ab = 1 \times 1 = 1.$

Thus, the answer is: $\boxed{1 \text{ (D)}}$

Would you like a more detailed explanation or have any questions?

Related Questions:

1. How can trigonometric values be estimated more precisely?
2. What are some properties of square roots in equations?
3. How do we verify solutions for trigonometric equations?
4. What are some efficient ways to solve integer constraint problems in equations?
5. Can this problem be generalized to other trigonometric functions?

Tip:

Remember that using known trigonometric approximations can speed up calculations when solving problems.