The user wants me to solve a multiple-choice question with multiple correct answers from the topic of inverse trigonometric functions.

Question Analysis: I'm given two angles, $\alpha$ and $\beta$, defined using inverse trigonometric functions. $\alpha = 3 \sin^{-1} \left( \frac{6}{11} \right)$ $\beta = 3 \cos^{-1} \left( \frac{4}{9} \right)$

The inverse trigonometric functions take only their principal values.

• For $\sin^{-1}(x)$, the range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
• For $\cos^{-1}(x)$, the range is $[0, \pi]$.

I need to determine the signs of $\cos \beta$, $\sin \beta$, $\cos(\alpha + \beta)$, and $\cos \alpha$.

Step 1: Analyze $\alpha = 3 \sin^{-1} \left( \frac{6}{11} \right)$

Let $\theta = \sin^{-1} \left( \frac{6}{11} \right)$. So, $\sin \theta = \frac{6}{11}$. Since $0 < \frac{6}{11} < 1$, we know that $0 < \theta < \frac{\pi}{2}$.

We need to find the range of $\alpha = 3\theta$. To do this, I need to compare $\frac{6}{11}$ with some standard values like $\frac{1}{2}$, $\frac{1}{\sqrt{2}}$, $\frac{\sqrt{3}}{2}$. $\sin(\frac{\pi}{6}) = \frac{1}{2} = 0.5$. $\frac{6}{11} \approx 0.545$. So, $\frac{6}{11} > \frac{1}{2}$. This means $\sin \theta > \sin(\frac{\pi}{6})$, which implies $\theta > \frac{\pi}{6}$ (since sine is increasing in $[0, \frac{\pi}{2}]$).

Let's check another value. $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \approx 0.707$. $\frac{6}{11} \approx 0.545 < 0.707$. So, $\sin \theta < \sin(\frac{\pi}{4})$, which implies $\theta < \frac{\pi}{4}$.

So we have $\frac{\pi}{6} < \theta < \frac{\pi}{4}$. Therefore, $3 \times \frac{\pi}{6} < 3\theta < 3 \times \frac{\pi}{4}$. $\frac{\pi}{2} < \alpha < \frac{3\pi}{4}$.

This tells us that $\alpha$ is in the second quadrant.

Now let's evaluate option D: $\cos \alpha < 0$. Since $\alpha$ is in the second quadrant ($\frac{\pi}{2} < \alpha < \frac{3\pi}{4}$), its cosine must be negative. So, $\cos \alpha < 0$. Option D is correct.

Step 2: Analyze $\beta = 3 \cos^{-1} \left( \frac{4}{9} \right)$

Let $\phi = \cos^{-1} \left( \frac{4}{9} \right)$. So, $\cos \phi = \frac{4}{9}$. Since $0 < \frac{4}{9} < 1$, we know that $0 < \phi < \frac{\pi}{2}$.

We need to find the range of $\beta = 3\phi$. Let's compare $\frac{4}{9}$ with standard values. $\cos(\frac{\pi}{3}) = \frac{1}{2} = 0.5$. $\frac{4}{9} \approx 0.444$. So, $\frac{4}{9} < \frac{1}{2}$. This means $\cos \phi < \cos(\frac{\pi}{3})$. Since cosine is a decreasing function in $[0, \frac{\pi}{2}]$, this implies $\phi > \frac{\pi}{3}$.

Let's check another value. $\cos(\frac{\pi}{2}) = 0$. $\frac{4}{9} > 0$, so $\phi < \frac{\pi}{2}$.

Let's check $\cos(\frac{2\pi}{5}) = \frac{\sqrt{5}-1}{4} \approx \frac{2.236-1}{4} \approx 0.309$. $\frac{4}{9} > \frac{\sqrt{5}-1}{4}$, so $\phi < \frac{2\pi}{5}$.

We have $\frac{\pi}{3} < \phi < \frac{\pi}{2}$. Therefore, $3 \times \frac{\pi}{3} < 3\phi < 3 \times \frac{\pi}{2}$. $\pi < \beta < \frac{3\pi}{2}$.

This tells us that $\beta$ is in the third quadrant.

Now let's evaluate option A: $\cos \beta > 0$. Since $\beta$ is in the third quadrant ($\pi < \beta < \frac{3\pi}{2}$), its cosine must be negative. So, $\cos \beta < 0$. Option A is incorrect.

Now let's evaluate option B: $\sin \beta < 0$. Since $\beta$ is in the third quadrant ($\pi < \beta < \frac{3\pi}{2}$), its sine must be negative. So, $\sin \beta < 0$. Option B is correct.

Step 3: Analyze $\cos(\alpha + \beta)$

We have the ranges for $\alpha$ and $\beta$: $\frac{\pi}{2} < \alpha < \frac{3\pi}{4}$ $\pi < \beta < \frac{3\pi}{2}$

Let's find the range of $\alpha + \beta$. Adding the inequalities: $\frac{\pi}{2} + \pi < \alpha + \beta < \frac{3\pi}{4} + \frac{3\pi}{2}$ $\frac{3\pi}{2} < \alpha + \beta < \frac{3\pi + 6\pi}{4} = \frac{9\pi}{4}$

So, $\alpha + \beta$ lies in the range $(\frac{3\pi}{2}, \frac{9\pi}{4})$. Let's analyze this range. $\frac{3\pi}{2}$ is 270 degrees. $\frac{9\pi}{4} = 2\pi + \frac{\pi}{4}$ is 360 + 45 = 405 degrees. The interval $(\frac{3\pi}{2}, \frac{9\pi}{4})$ covers the entire fourth quadrant $(\frac{3\pi}{2}, 2\pi)$ and part of the first quadrant $(2\pi, \frac{9\pi}{4})$. In the fourth quadrant, cosine is positive. In the first quadrant, cosine is positive. So, for any value of $\alpha + \beta$ in this range, $\cos(\alpha + \beta)$ will be positive.

Therefore, $\cos(\alpha + \beta) > 0$. Option C is correct.

Summary of findings:

• Option A: $\cos\beta > 0$. We found $\pi < \beta < \frac{3\pi}{2}$, so $\cos \beta < 0$. Incorrect.
• Option B: $\sin \beta < 0$. We found $\pi < \beta < \frac{3\pi}{2}$, so $\sin \beta < 0$. Correct.
• Option C: $\cos(\alpha + \beta) > 0$. We found $\frac{3\pi}{2} < \alpha + \beta < \frac{9\pi}{4}$, so $\cos(\alpha + \beta) > 0$. Correct.
• Option D: $\cos \alpha < 0$. We found $\frac{\pi}{2} < \alpha < \frac{3\pi}{4}$, so $\cos \alpha < 0$. Correct.

The correct options are B, C, and D. This matches the stored correct answer.