Question

Let tan A, tan B, where A, B $\in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, be the roots of the quadratic equation $x^2 - 2x - 5 = 0$. Then $20 \sin^2\left(\frac{A+B}{2}\right)$ is equal to:

• A. $10+\sqrt{10}$
• B. $10-2\sqrt{10}$
• C. $10-3\sqrt{10}$
• D. $10-\sqrt{10}$

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The roots of a given quadratic equation are trigonometric functions, $\tan A$ and $\tan B$. We need to evaluate an expression involving the angle $\frac{A+B}{2}$. The problem connects the theory of quadratic equations with trigonometric identities.
Step 2: Key Formula or Approach:
1. Vieta's Formulas: Find the sum ($\tan A + \tan B$) and product ($\tan A \tan B$) of the roots from the quadratic equation.
2. Tangent Addition Formula: Use the sum and product of the roots to find $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.
3. Pythagorean Identity: Find $\cos(A+B)$ from $\tan(A+B)$. We will need to determine the correct sign of the cosine.
4. Half-Angle Identity: Use the identity $2\sin^2(\theta) = 1 - \cos(2\theta)$ with $\theta = \frac{A+B}{2}$ to relate the target expression to $\cos(A+B)$.
Step 3: Detailed Explanation:
For the quadratic equation $x^2 - 2x - 5 = 0$ with roots $\tan A$ and $\tan B$:
By Vieta's formulas:
Sum of roots: $\tan A + \tan B = 2$
Product of roots: $\tan A \tan B = -5$
Using the tangent addition formula, we find $\tan(A+B)$:
\[ \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{2}{1 - (-5)} = \frac{2}{6} = \frac{1}{3} \] Now, we find $\cos(A+B)$. We know that $\sec^2(X) = 1 + \tan^2(X)$.
\[ \sec^2(A+B) = 1 + \left(\frac{1}{3}\right)^2 = 1 + \frac{1}{9} = \frac{10}{9} \] Therefore, $\cos^2(A+B) = \frac{1}{\sec^2(A+B)} = \frac{9}{10}$, which implies $\cos(A+B) = \pm\frac{3}{\sqrt{10}}$.
To determine the sign, let's find the quadrant of $A+B$. Given $A, B \in (-\frac{\pi}{2}, \frac{\pi}{2})$. The product $\tan A \tan B = -5$ is negative, so one angle is in $(-\frac{\pi}{2}, 0)$ and the other is in $(0, \frac{\pi}{2})$. The sum $\tan A + \tan B = 2$ is positive, meaning the positive tangent value is larger in magnitude. Therefore, the sum $A+B$ must lie in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$. Since $\tan(A+B) = 1/3$ is positive, $A+B$ must be in the first quadrant, so $\cos(A+B)$ is positive.
\[ \cos(A+B) = \frac{3}{\sqrt{10}} \] The expression we need to evaluate is $20 \sin^2\left(\frac{A+B}{2}\right)$. Let's use the half-angle identity:
\[ 20 \sin^2\left(\frac{A+B}{2}\right) = 10 \left[2\sin^2\left(\frac{A+B}{2}\right)\right] \] \[ = 10 \left[1 - \cos\left(2 \cdot \frac{A+B}{2}\right)\right] = 10 [1 - \cos(A+B)] \] Substitute the value of $\cos(A+B)$:
\[ = 10 \left(1 - \frac{3}{\sqrt{10}}\right) = 10 - \frac{30}{\sqrt{10}} \] Rationalizing the second term:
\[ = 10 - \frac{30\sqrt{10}}{10} = 10 - 3\sqrt{10} \] Step 4: Final Answer:
The value of $20 \sin^2\left(\frac{A+B}{2}\right)$ is $10-3\sqrt{10}$.