Question

If A and B are positive acute angles satisfying $3\cos^2 A + 2\cos^2 B = 4$ and $\frac{3\sin A{\sin B} = \frac{2\cos B}{\cos A}$, then A+2B=}

• A. \( 30^\circ \)
• B. \( 45^\circ \)
• C. \( 60^\circ \)
• D. \( 90^\circ \)

Hint:

For trigonometric problems involving multiple equations and unknown angles, especially when the required answer is a sum/difference of angles (e.g., $A+2B$), and the options are standard angles ($30^\circ, 45^\circ, 60^\circ, 90^\circ$), it's often effective to: % Option (A) Simplify the given equations using standard trigonometric identities (e.g., double angle formulas, $\sin^2\theta + \cos^2\theta = 1$). % Option (B) If direct calculation of $A$ and $B$ is difficult, assume the result (e.g., $A+2B = 90^\circ$) and substitute it back into the original equations or derived simplified equations. If it leads to consistency, that's likely the correct answer. This technique is particularly useful in multiple-choice questions.

Answer 1

The Correct Option is D

We are given two equations:
1) \( 3\cos^2 A + 2\cos^2 B = 4 \)
2) \( \frac{3\sin A}{\sin B} = \frac{2\cos B}{\cos A} \)

Step 1: Manipulating the second equation

From the second equation, cross-multiply: \[ 3\sin A \cos A = 2\sin B \cos B \] Using the double angle identity \( \sin(2\theta) = 2\sin\theta\cos\theta \), we get: \[ \frac{3}{2}(2\sin A \cos A) = (2\sin B \cos B) \] \[ \frac{3}{2}\sin(2A) = \sin(2B) \] \[ 3\sin(2A) = 2\sin(2B) \quad \text{(Equation X)} \]

Step 2: Manipulating the first equation

Using the identity \( \cos^2\theta = \frac{1+\cos(2\theta)}{2} \), we substitute into the first equation: \[ 3\left(\frac{1+\cos(2A)}{2}\right) + 2\left(\frac{1+\cos(2B)}{2}\right) = 4 \] Multiply by 2: \[ 3(1+\cos(2A)) + 2(1+\cos(2B)) = 8 \] \[ 3 + 3\cos(2A) + 2 + 2\cos(2B) = 8 \] \[ 5 + 3\cos(2A) + 2\cos(2B) = 8 \] \[ 3\cos(2A) + 2\cos(2B) = 3 \quad \text{(Equation Y)} \]

Step 3: Solving the system of equations

From Equation X: \[ \sin(2B) = \frac{3}{2}\sin(2A) \] From Equation Y: \[ 2\cos(2B) = 3 - 3\cos(2A) \implies \cos(2B) = \frac{3 - 3\cos(2A)}{2} \]

Step 4: Using the identity \( \sin^2(2B) + \cos^2(2B) = 1 \)

Substitute into the identity: \[ \left(\frac{3}{2}\sin(2A)\right)^2 + \left(\frac{3 - 3\cos(2A)}{2}\right)^2 = 1 \] Multiply by 4: \[ 9\sin^2(2A) + 9(1 - 2\cos(2A) + \cos^2(2A)) = 4 \] \[ 9(\sin^2(2A) + \cos^2(2A)) + 9 - 18\cos(2A) = 4 \] \[ 9(1) + 9 - 18\cos(2A) = 4 \] \[ 18 - 18\cos(2A) = 4 \] \[ 18\cos(2A) = 14 \implies \cos(2A) = \frac{7}{9} \]

Step 5: Finding \( \cos(2B) \)

Substitute \( \cos(2A) = \frac{7}{9} \) into the expression for \( \cos(2B) \): \[ \cos(2B) = \frac{3 - 3\left(\frac{7}{9}\right)}{2} = \frac{3 - \frac{7}{3}}{2} = \frac{\frac{9-7}{3}}{2} = \frac{2/3}{2} = \frac{1}{3} \]

Step 6: Checking \( A + 2B = 90^\circ \)

We need to test if \( A + 2B = 90^\circ \). If \( A + 2B = 90^\circ \), then \( A = 90^\circ - 2B \). Substituting this into the second equation: \[ \frac{3\sin(90^\circ - 2B)}{\sin B} = \frac{2\cos B}{\cos(90^\circ - 2B)} \] \[ \frac{3\cos(2B)}{\sin B} = \frac{2\cos B}{\sin(2B)} \] Cross-multiply: \[ 3\cos(2B)\sin(2B) = 2\sin B \cos B \] Using the identity \( 2\sin\theta\cos\theta = \sin(2\theta) \), we get: \[ 3\cos(2B)\sin(2B) = \sin(2B) \] Dividing by \( \sin(2B) \): \[ 3\cos(2B) = 1 \] \[ \cos(2B) = \frac{1}{3} \] This matches the value we derived for \( \cos(2B) \). Since \( A \) and \( B \) are acute angles, this confirms \( A + 2B = 90^\circ \). Therefore, \( A + 2B = 90^\circ \) is true.