Question
Let $\alpha = 3\sin^{- 1}\left( \frac{6}{11} \right)$ and $\beta = 3\cos^{- 1}\left( \frac{4}{9} \right)$, where inverse trigonometric functions take only the principal values.
Given below are two statements:
Statement I : $cos(\alpha + \beta) > 0$
Statement II : $cos(\alpha) < 0$
In the light of the above statements, choose the correct answer from the options given below:
Options
Solution
Sol. [Key Concept: Bound $\alpha = 3\sin^{- 1}\frac{6}{11} \in (\frac{\pi}{2,}\frac{3\pi}{4)}$ and $\beta = 3\cos^{- 1}\frac{4}{9} \in (\pi,\frac{3\pi}{2)}$: then $cos\alpha < 0$ and $cos(\alpha + \beta) > 0$, so both statements are true.]
$$\frac{1}{2} < \frac{6}{11} < \frac{1}{\sqrt{2}}$$
$$\begin{matrix} & \sin^{- 1}\left( \frac{1}{2} \right) < \sin^{- 1}\left( \frac{6}{11} \right) < \sin^{- 1}\left( \frac{1}{\sqrt{2}} \right) \\ & \frac{\pi}{6} < 3\sin^{- 1}\left( \frac{6}{11} \right) < \frac{\pi}{4} \\ & \frac{\pi}{2} < \alpha < \frac{3\pi}{4}\ \therefore cos\alpha < 0 \\ & 0 < \frac{4}{9} < \frac{1}{2} \\ & \frac{\pi}{3} < \cos^{- 1}\left( \frac{4}{9} \right) < \frac{\pi}{2} \\ & \pi < 3\cos^{- 1}\left( \frac{4}{9} \right) < \frac{3\pi}{2} \\ & \pi < \beta < \frac{3\pi}{2} \end{matrix}$$
Now $\frac{3\pi}{2} < \alpha + \beta < \frac{9\pi}{4}$
$$\therefore cos(\alpha + \beta) > 0$$
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