Question

Let $\alpha =3{\sin }^{-1}\left(\frac{6}{11}\right)$ and $\beta =3{\cos }^{-1}\left(\frac{4}{9}\right)$ where the inverse trigonometric functions take only the principal values. Then

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

Answer 1

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

$\cos \alpha <0$

$\alpha =3{\sin }^{-1}\frac{6}{11},\beta =3{\cos }^{-1}\frac{4}{9}$
Since $3{\sin }^{-1}\frac{6}{11}>3{\sin }^{-1}\frac{1}{2}=\frac{\pi }{2},$
$\therefore \alpha >\frac{\pi }{2}$
$\therefore \cos \alpha <0$ and $\sin \alpha >0$

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