Let =3^-1(611) and =3^-1(49), where inverse trigonometric functions take only the principa…

Question

Let $\alpha=3\sin^{-1}\left(\dfrac{6}{11}\right)$ and $\beta=3\cos^{-1}\left(\dfrac49\right)$, where inverse trigonometric functions take only the principal values.

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 :

Accepted Answer

Since $\dfrac12<\dfrac{6}{11}<\dfrac{1}{\sqrt2}$, $$\sin^{-1}\frac12<\sin^{-1}\frac{6}{11}<\sin^{-1}\frac{1}{\sqrt2}\ \Rightarrow\ \frac{\pi}{2}<\alpha<\frac{3\pi}{4}\ \Rightarrow\ \cos\alpha<0.\quad\text{(Statement II is true)}$$ Also $\dfrac49<\dfrac12<\dfrac{1}{\sqrt2}$, so $\cos^{-1}\dfrac49>\dfrac{\pi}{3}>\dfrac{\pi}{4}$, giving $\beta=3\cos^{-1}\dfrac49>\pi$. Since $\alpha>\dfrac{\pi}{2}$ and $\beta>\pi$, $\alpha+\beta>\dfrac{3\pi}{2}$, which lies where $\cos(\alpha+\beta)>0$ is consistent (the two inequalities together place $\alpha+\beta$ in a range giving a positive cosine). Hence Statement I is also true. Both statements are true.

Solutions · 1