Question

If \(\alpha = 3\sin^{-1}\left(\frac{6}{11}\right)\) and \(\beta = 3\cos^{-1}\left(\frac{4}{9}\right)\), consider statements:
Statement 1: \(\cos(\alpha + \beta) > 0\)
Statement 2: \(\cos\alpha < 0\)
Then which of the following is true?

• A. Statement 1 and 2 are correct
• B. Only statement 1 is correct
• C. Only statement 2 is correct
• D. None of these statements are correct

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To determine the sign of the cosine values, we must locate the quadrants for $α$ and $β$. We approximate the inverse trigonometric values by comparing them with standard values like $\sin(30^\circ) = 0.5$ and $\sin(45^\circ) ≈ 0.707$.

Step 2: Key Formula or Approach:
1. For $α = 3\theta₁$, where $\theta₁ = \sin^-1(6/11)$. 2. For $β = 3\theta₂$, where $\theta₂ = \cos^-1(4/9)$.

Step 3: Detailed Explanation:
1. **Evaluating \(\alpha\):** - \( \sin 30^\circ = 0.5 \) and \( \sin 45^\circ \approx 0.707 \). Since \( 0.5 < 6/11 < 0.707 \), \( 30^\circ < \theta_1 < 45^\circ \). - Multiplying by 3: \( 90^\circ < 3\theta_1 < 135^\circ \). - Thus, \(\alpha\) is in the 2nd quadrant. In the 2nd quadrant, \(\cos \alpha < 0\). **Statement 2 is correct.** 2. **Evaluating \(\beta\):** - \( \cos 60^\circ = 0.5 \) and \( \cos 90^\circ = 0 \). Since \( 0 < 4/9 < 0.5 \), \( 60^\circ < \theta_2 < 90^\circ \). - Multiplying by 3: \( 180^\circ < 3\theta_2 < 270^\circ \). - Thus, \(\beta\) is in the 3rd quadrant. 3. **Evaluating \(\cos(\alpha + \beta)\):** - \( \alpha + \beta \) range: \( (90^\circ + 180^\circ) < \alpha + \beta < (135^\circ + 270^\circ) \). - \( 270^\circ < \alpha + \beta < 405^\circ \). - This spans the 4th quadrant and the beginning of the 1st quadrant. In both these regions, cosine is positive. - Therefore, \(\cos(\alpha + \beta) > 0\). **Statement 1 is correct.**

Step 4: Final Answer:
Both Statement 1 and 2 are correct.