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:
We need to determine the range of $\alpha$ and $\beta$ by bounding the argument of each inverse trigonometric function, and then deduce the signs of $\cos\alpha$ and $\cos(\alpha+\beta)$.
The key idea is to use the monotonicity of $\sin^{-1}$ and $\cos^{-1}$: since $\sin^{-1}$ is increasing and $\cos^{-1}$ is decreasing, bounding the argument gives us a bound on the angle.
Step 2: Finding the range of $\alpha$:
We observe that: \[ \frac{1}{2}<\frac{6}{11}<\frac{1}{\sqrt{2}} \] Applying $\sin^{-1}$ (which is an increasing function) to all three parts: \[ \sin^{-1}\!\left(\frac{1}{2}\right)<\sin^{-1}\!\left(\frac{6}{11}\right)<\sin^{-1}\!\left(\frac{1}{\sqrt{2}}\right) \] \[ \frac{\pi}{6}<\sin^{-1}\!\left(\frac{6}{11}\right)<\frac{\pi}{4} \] Multiplying throughout by 3: \[ \frac{\pi}{2}<\alpha = 3\sin^{-1}\!\left(\frac{6}{11}\right)<\frac{3\pi}{4} \] Since $\alpha \in \left(\dfrac{\pi}{2},\, \dfrac{3\pi}{4}\right)$, we conclude that $\cos\alpha<0$.
Therefore, Statement 2 is TRUE.
Step 3: Finding the range of $\beta$:
We observe that: \[ 0<\frac{4}{9}<\frac{1}{2} \] Applying $\cos^{-1}$ (which is a decreasing function): \[ \cos^{-1}\!\left(\frac{1}{2}\right)<\cos^{-1}\!\left(\frac{4}{9}\right)<\cos^{-1}(0) \] \[ \frac{\pi}{3}<\cos^{-1}\!\left(\frac{4}{9}\right)<\frac{\pi}{2} \] Multiplying throughout by 3: \[ \pi<\beta = 3\cos^{-1}\!\left(\frac{4}{9}\right)<\frac{3\pi}{2} \] Step 4: Finding the range of $\alpha + \beta$:
Adding the two ranges: \[ \frac{\pi}{2} + \pi<\alpha + \beta<\frac{3\pi}{4} + \frac{3\pi}{2} \] \[ \frac{3\pi}{2}<\alpha + \beta<\frac{9\pi}{4} \] The interval $\left(\dfrac{3\pi}{2},\, \dfrac{9\pi}{4}\right)$ lies in the fourth quadrant region (between $\dfrac{3\pi}{2}$ and $2\pi$) and beyond $2\pi$ into the first quadrant.
In both these sub-regions, cosine is positive (since $\cos\theta>0$ for $\theta \in \left(\dfrac{3\pi}{2}, 2\pi\right)$ and $\theta \in \left(2\pi, \dfrac{9\pi}{4}\right)$).
Therefore $\cos(\alpha + \beta)>0$.
Statement 1 is TRUE.
Step 5: Final Answer:
Both Statement 1 ($\cos(\alpha+\beta)>0$) and Statement 2 ($\cos\alpha<0$) are correct.
The answer is Option (1).