Question

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:

• A. Both Statement I and Statement II are true
• B. Both Statement I and Statement II are false
• C. Statement I is true but Statement II is false
• D. Statement I is false but Statement II is true

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We must estimate the approximate values of \( \alpha \) and \( \beta \) by bounding the arguments of the inverse trigonometric functions using standard known values. This will help determine the quadrants in which \( \alpha \) and \(\alpha+\beta\) lie, dictating the sign of their cosine values.
Step 2: Key Formula or Approach:
Use bounding values:
For sine: \( \sin(\pi/6) = 0.5 \), \( \sin(\pi/4) \approx 0.707 \), \( \sin(\pi/3) \approx 0.866 \).
For cosine: \( \cos(\pi/3) = 0.5 \), \( \cos(\pi/2) = 0 \).
Determine ranges for \( \alpha/3 \) and \( \beta/3 \) and subsequently ranges for \( \alpha \) and \( \beta \).
Step 3: Detailed Explanation:
Let's analyze \( \alpha = 3 \sin^{-1} \left( \frac{6}{11} \right) \).
\( \frac{6}{11} \approx 0.545 \).
We know \( \sin(\frac{\pi}{6}) = \frac{1}{2} = 0.5 \) and \( \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \approx 0.707 \).
Since \( 0.5<0.545<0.707 \), we have \( \frac{\pi}{6}<\sin^{-1} \left( \frac{6}{11} \right)<\frac{\pi}{4} \).
Multiply by 3 to find the range of \( \alpha \):
\( 3 \left( \frac{\pi}{6} \right)<\alpha<3 \left( \frac{\pi}{4} \right) \implies \frac{\pi}{2}<\alpha<\frac{3\pi}{4} \).
Since \( \alpha \) is in the second quadrant, \( \cos \alpha \) must be negative.
Thus, \( \cos \alpha<0 \). Statement II is true.
Now let's analyze \( \beta = 3 \cos^{-1} \left( \frac{4}{9} \right) \).
\( \frac{4}{9} \approx 0.444 \).
We know \( \cos(\frac{\pi}{3}) = \frac{1}{2} = 0.5 \) and \( \cos(\frac{\pi}{2}) = 0 \).
Since cosine is decreasing in the first quadrant and \( 0<0.444<0.5 \), we have:
\( \frac{\pi}{3}<\cos^{-1} \left( \frac{4}{9} \right)<\frac{\pi}{2} \).
Multiply by 3 to find the range of \( \beta \):
\( 3 \left( \frac{\pi}{3} \right)<\beta<3 \left( \frac{\pi}{2} \right) \implies \pi<\beta<\frac{3\pi}{2} \).
Now, let's find the range of \( \alpha + \beta \):
Add the ranges of \( \alpha \) and \( \beta \):
\( \frac{\pi}{2} + \pi<\alpha + \beta<\frac{3\pi}{4} + \frac{3\pi}{2} \)
\( \frac{3\pi}{2}<\alpha + \beta<\frac{9\pi}{4} \).
The angle \( \alpha + \beta \) falls between \( 270^\circ \) and \( 405^\circ \). This means it is in the fourth quadrant (or just crosses into the first).
Wait, let's look closer. Is it definitely in the fourth quadrant or could it be in the first?
Let's approximate better.
\( \sin(\alpha/3) \approx 0.545 \implies \alpha/3 \approx 33^\circ \implies \alpha \approx 99^\circ \).
\( \cos(\beta/3) \approx 0.444 \implies \beta/3 \approx 63.6^\circ \implies \beta \approx 190.8^\circ \).
Sum: \( \alpha + \beta \approx 99^\circ + 190.8^\circ = 289.8^\circ \).
An angle of \( 289.8^\circ \) is strictly in the fourth quadrant (\( 270^\circ<289.8^\circ<360^\circ \)).
In the fourth quadrant, the cosine function is positive.
Thus, \( \cos(\alpha + \beta)>0 \). Statement I is true.
Step 4: Final Answer:
Both Statement I and Statement II are true.