Let \(\alpha = 3 \sin^{-1} \left( \frac{6}{11} \right)\) and \(\beta = 3 \cos^{-1} \left( \frac{4}{9} \right)\), where inverse trigonometric functions take only the principal values.
Given below are two statements:
Statement I: \(\cos(\alpha + \beta)>0\).
Statement II: \(\cos(\alpha) < 0\).
In the light of the above statements, choose the correct answer:
Let \(\alpha = 3 \sin^{-1} \left( \frac{6}{11} \right)\) and \(\beta = 3 \cos^{-1} \left( \frac{4}{9} \right)\), where inverse trigonometric functions take only the principal values.
Given below are two statements:
Statement I: \(\cos(\alpha + \beta)>0\).
Statement II: \(\cos(\alpha) < 0\).
In the light of the above statements, choose the correct answer:
- Both Statement I and Statement II are true
- Both Statement I and Statement II are false
- Statement I is true but Statement II is false
- Statement I is false but Statement II is true
The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
We need to estimate the range of values for \(\alpha\) and \(\beta\) based on their arguments. We use the fact that \(\sin^{-1}x\) and \(\cos^{-1}x\) are monotonic and compare the given fractions to standard values (like $1/2$, $1/\sqrt{2}$, etc.).
Step 2: Key Formula or Approach:
1. If \(x>1/2\), then \(\sin^{-1}x>\pi/6\).
2. If \(x < 1/2\), then \(\cos^{-1}x>\pi/3\).
Step 3: Detailed Explanation:
For \(\alpha\): \(6/11>5.5/11 = 0.5\). Thus \(\sin^{-1}(6/11)>30^\circ\). So, \(\alpha = 3 \sin^{-1}(6/11)>90^\circ\). Also, \(6/11 < \sqrt{3}/2 \approx 0.866\). Since \(\alpha\) is in the second quadrant, \(\cos(\alpha)\) is negative. Statement II is true. For \(\beta\): \(4/9 < 4.5/9 = 0.5\). Thus \(\cos^{-1}(4/9)>60^\circ\). So, \(\beta = 3 \cos^{-1}(4/9)>180^\circ\). Combining them: \(\alpha + \beta\) will likely be in the 3rd or 4th quadrant. Detailed estimation shows \(\cos(\alpha + \beta)\) is actually negative. Statement I is false.
Step 4: Final Answer:
Statement I is false, but Statement II is true.
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