Evaluate Sign of Trigonometric Functions Involving Inverse Angles

To determine the correct statement(s), we analyze the values and ranges of the angles $\alpha$ and $\beta$ step-by-step.

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Step 1: Analysis of Angle $\alpha$

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

Let $\theta_1 = \sin^{-1}\left(\frac{6}{11}\right)$. Since $\sin x$ is a strictly increasing function on $\left[0, \frac{\pi}{2}\right]$: $\frac{1}{2} < \frac{6}{11} < \frac{\sqrt{3}}{2}$

Taking the inverse sine across the inequality: $\sin^{-1}\left(\frac{1}{2}\right) < \sin^{-1}\left(\frac{6}{11}\right) < \sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$ $\frac{\pi}{6} < \theta_1 < \frac{\pi}{3}$

Multiplying the entire inequality by $3$: $\frac{\pi}{2} < 3\theta_1 < \pi \implies \frac{\pi}{2} < \alpha < \pi$

Since $\alpha$ lies in the second quadrant ($\text{Quadrant II}$):

• $\cos\alpha < 0$

Thus, Option (D) is correct.

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Step 2: Analysis of Angle $\beta$

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

Let $\theta_2 = \cos^{-1}\left(\frac{4}{9}\right)$. Since $\cos x$ is a strictly decreasing function on $\left[0, \frac{\pi}{2}\right]$: $0 < \frac{4}{9} < \frac{1}{2}$

Taking the inverse cosine across the inequality: $\cos^{-1}\left(\frac{1}{2}\right) < \cos^{-1}\left(\frac{4}{9}\right) < \cos^{-1}(0)$ $\frac{\pi}{3} < \theta_2 < \frac{\pi}{2}$

Multiplying the entire inequality by $3$: $\pi < 3\theta_2 < \frac{3\pi}{2} \implies \pi < \beta < \frac{3\pi}{2}$

Since $\beta$ lies in the third quadrant ($\text{Quadrant III}$):

• $\cos\beta < 0$ (making Option (A) incorrect)
• $\sin\beta < 0$

Thus, Option (B) is correct.

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Step 3: Analysis of Angle $(\alpha + \beta)$

We can express $\alpha + \beta$ in terms of $\theta_1$ and $\theta_2$: $\alpha + \beta = 3(\theta_1 + \theta_2)$

Since $\theta_2 = \frac{\pi}{2} - \sin^{-1}\left(\frac{4}{9}\right)$, we have: $\theta_1 + \theta_2 = \frac{\pi}{2} + \left(\sin^{-1}\left(\frac{6}{11}\right) - \sin^{-1}\left(\frac{4}{9}\right)\right)$

Let $\delta = \sin^{-1}\left(\frac{6}{11}\right) - \sin^{-1}\left(\frac{4}{9}\right)$. Using the sine subtraction formula $\sin(A - B) = \sin A \cos B - \cos A \sin B$: $\sin\delta = \frac{6}{11}\sqrt{1 - \left(\frac{4}{9}\right)^2} - \frac{4}{9}\sqrt{1 - \left(\frac{6}{11}\right)^2} = \frac{6\sqrt{65} - 4\sqrt{85}}{99}$

Approximating the radicals:

• $6\sqrt{65} \approx 6 \times 8.062 = 48.372$
• $4\sqrt{85} \approx 4 \times 9.220 = 36.880$

Thus: $\sin\delta \approx \frac{48.372 - 36.880}{99} = \frac{11.492}{99} \approx 0.116 < \frac{1}{2} = \sin\left(\frac{\pi}{6}\right)$

Since $0 < \sin\delta < \sin\left(\frac{\pi}{6}\right)$, we get $0 < \delta < \frac{\pi}{6}$.

Substituting this back into the expression for $\theta_1 + \theta_2$: $\frac{\pi}{2} < \theta_1 + \theta_2 < \frac{\pi}{2} + \frac{\pi}{6} = \frac{2\pi}{3}$

Multiplying through by $3$: $\frac{3\pi}{2} < 3(\theta_1 + \theta_2) < 2\pi \implies \frac{3\pi}{2} < \alpha + \beta < 2\pi$

This shows that $(\alpha + \beta)$ lies strictly in the fourth quadrant ($\text{Quadrant IV}$). In this quadrant, the cosine function is positive:

• $\cos(\alpha + \beta) > 0$

Thus, Option (C) is correct.

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Conclusion

The correct options are B, C, and D.