Inverse Circular Functions 2 Question 4
4. If $\alpha=3 \sin ^{-1} \frac{6}{11}$ and $\beta=3 \cos ^{-1} \frac{4}{9}$, where the inverse trigonometric functions take only the principal values, then the correct option(s) is/are
(2015 Adv.)
(a) $\cos \beta>0$
(b) $\sin \beta<0$
(c) $\cos (\alpha+\beta)>0$
(d) $\cos \alpha<0$
Show Answer
Answer:
Correct Answer: 4. (c)
Solution:
- Here, $\alpha=3 \sin ^{-1} \frac{6}{11}$ and $\beta=3 \cos ^{-1} \frac{4}{9}$ as $\frac{6}{11}>\frac{1}{2}$
$$ \begin{aligned} & \Rightarrow \quad \sin ^{-1} \frac{6}{11}>\sin ^{-1} \frac{1}{2}=\frac{\pi}{6} \\ & \therefore \quad \alpha=3 \sin ^{-1} \frac{6}{11}>\frac{\pi}{2} \\ & \Rightarrow \quad \cos \alpha<0 \end{aligned} $$
Now, $\quad \beta=3 \cos ^{-1} \frac{4}{9}$
As $\quad \frac{4}{9}<\frac{1}{2} \Rightarrow \cos ^{-1} \frac{4}{9}>\cos ^{-1} \frac{1}{2}=\frac{\pi}{3}$
$\therefore \quad \beta=3 \cos ^{-1} \frac{4}{9}>\pi$
$\therefore \quad \cos \beta<0$ and $\sin \beta<0$
Now, $\alpha+\beta$ is slightly greater than $\frac{3 \pi}{2}$.
$\therefore \quad \cos (\alpha+\beta)>0$
