Let =3 ^-1((6)/(11)) and =3 ^-1((4)/(9)), where inverse trigonometric functions take only …

Question

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

Given below are two statements :

Statement I : $\, \cos (\alpha +\beta )>0$

Accepted Answer

$\frac{1}{2}$\Rightarrow 3{\sin }\^{-1}(\frac{1}{2})$\frac{4}{9}\frac{\pi }{3}>\frac{\pi }{4}$

$\Rightarrow 3{\cos }^{-1}(\frac{4}{9})>\pi >\frac{3\pi }{4}\Rightarrow \beta >\pi \ldots \ldots (ii)$

Inequalities $(i)$ and $(ii)$ shows the mathematical logic for $\alpha +\beta >\frac{\pi }{2}+\pi $

$\Rightarrow \cos (\alpha +\beta )>0\ldots \ldots$ (iii)

From equations (i) and (iii) we can conclude :

$\begin{gathered} \ \end{gathered}$

Solutions · 1