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

Answer 1

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

$\sin \beta <0$

$\alpha =3{\sin }^{-1}\frac{6}{11}$
since $\frac{\pi }{4}>{\sin }^{-1}\frac{6}{11}>\frac{\pi }{6}$
$\frac{3\pi }{4}>3{\sin }^{-1}\frac{6}{11}>\frac{\pi }{2}$
$\cos \alpha <0$
$\cos \left(3{\cos }^{-1}\frac{4}{9}\right)=4\times {\left(\frac{4}{9}\right)}^3-3\times \frac{4}{9}$
$=\frac{4\times 64}{729}\frac{12}{9}$
= $\frac{256-12\times 81}{72}<0$
$\frac{5\pi }{12}>{\cos }^{-1}\frac{4}{9}>\frac{\pi }{3}$
$\frac{5\pi }{4}>3{\cos }^{-1}\frac{4}{9}>\pi$
$\sin \left(3{\cos }^{-1}\frac{4}{9}\right)<0$
$\frac{3\pi }{4}>3{\sin }^{-1}\frac{6}{11}>\frac{\pi }{2}$
$\frac{5\pi }{4}>3{\cos }^{-1}\frac{4}{9}>\pi$
$2\pi >3{\sin }^{-1}\frac{6}{11}+3{\cos }^{-1}\frac{4}{9}>\frac{3\pi }{2}$
$\cos (3{\sin }^{-1}\frac{6}{11}+3{\cos }^{-1}\frac{4}{9})>0$