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)

$\cos \alpha <0$

Explanation for the correct option:

Step 1: Find the intervals in which$\mathit{\alpha }\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{\beta }$lies.

$$\begin{gathered} \alpha =3{\sin }^{-1}\left(\frac{6}{11}\right)>3{\sin }^{-1}\left(\frac{6}{12}\right)=\frac{\mathrm{\pi }}{2} \\ \therefore \frac{\mathrm{\pi }}{2}<\alpha <\mathrm{\pi }......\left(1\right) \\ \mathrm{\beta }=3{\cos }^{-1}\left(\frac{4}{9}\right)>3{\cos }^{-1}\left(\frac{4}{8}\right)=\mathrm{\pi } \\ \therefore \mathrm{\pi }<\mathrm{\beta }<\frac{3\mathrm{\pi }}{2}......\left(2\right) \end{gathered}$$

Step 2: Check signs of$\mathit{s}\mathit{i}\mathit{n}\mathbf{,}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathbf{}\mathit{\alpha }\mathbf{,}\mathbf{}\mathit{\beta }\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{\alpha }\mathbf{+}\mathit{\beta }$.

$$\begin{gathered} \Rightarrow \sin \left(\alpha \right)>0and\sin \left(\beta \right)<0 \\ \Rightarrow \cos \left(\alpha \right)<0and\cos \left(\beta \right)<0 \\ \frac{3\mathrm{\pi }}{2}<\alpha +\beta <\frac{5\mathrm{\pi }}{2}(4thor1stquadrant) \\ \therefore \cos \left(\alpha +\beta \right)>0 \end{gathered}$$

Hence, options$\left(B\right),\left(C\right)and\left(D\right)$are the correct answers.