Question

If $\alpha =3{\sin }^{-1}\left(\frac{6}{11}\right)$ and $\beta =3{\cos }^{-1}\left(\frac{4}{9}\right)$, where the inverse trigonometric functions take only the principal values, then the correct option(s) is/are

• A. $\cos \beta >0$
• B. $\sin \beta <0$
• C. $\cos (\alpha +\beta )>0$
• D. $\cos \alpha <0$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $\sin \beta <0$
C $\cos (\alpha +\beta )>0$
D $\cos \alpha <0$

Here, $\alpha =3{\sin }^{-1}\left(\frac{6}{11}\right)$ and $\beta =3{\cos }^{-1}\left(\frac{4}{9}\right)$
As, $\frac{6}{11}>\frac{1}{2}\Rightarrow {\sin }^{-1}\left(\frac{6}{11}\right)>{\sin }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{6}\therefore \alpha =3{\sin }^{-1}\left(\frac{6}{11}\right)>\frac{\pi }{2}\Rightarrow \cos \alpha <0$
Now,
$\beta =3{\cos }^{-1}\left(\frac{4}{9}\right)$
As, $\frac{4}{9}<\frac{1}{2}\Rightarrow {\cos }^{-1}\left(\frac{4}{9}\right)>{\cos }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}\therefore \beta =3{\cos }^{-1}\left(\frac{4}{9}\right)>\pi \therefore \cos \beta <0$ and $\sin \beta <0$
Now, $\alpha +\beta$ is slightly greater than $\frac{3\pi }{2}$
$\therefore \cos (\alpha +\beta )>0$