Animated Solution for Mathematics - Inverse Trigonometric Functions: 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)

Welcome, future engineer. Today, we are not just solving a problem; we are going on a detective mission. We have two mysterious angles, $\alpha$ and $\beta$, defined by inverse trigonometric functions.

They are hiding in the shadows of the unit circle, and our job is to drag them out into the light, determine their quadrants, and verify their properties. Many students look at $\alpha =3{\sin }^{-1}\left(\frac{6}{11}\right)$ and immediately panic, reaching for a calculator that they aren't allowed to use.

But you have a better tool: your brain and the logic of inequalities.

Let us start with $\alpha =3{\sin }^{-1}\left(\frac{6}{11}\right)$. We need to know where this angle lives. To find out, we compare the input $\frac{6}{11}$ to values we know by heart.

We know that $\frac{6}{11}$ is slightly larger than $\frac{6}{12}$, which is $\frac{1}{2}$. Since the function $f(x)={\sin }^{-1}(x)$ is strictly increasing, we can confidently say:

Now, multiply this entire inequality by $3$. We get $\alpha >3\times \frac{\pi }{6}=\frac{\pi }{2}$. This is a massive breakthrough!

We now know $\alpha$ is greater than $\frac{\pi }{2}$. But how far does it go? We also know that $\frac{6}{11}$ is less than $\frac{\sqrt{3}}{2}$ (which is approximately $0.866$).

Therefore, $\alpha <3\times \frac{\pi }{3}=\pi$. So, we have trapped $\alpha$ in the interval $\left(\frac{\pi }{2},\pi \right)$.

This is the second quadrant. In the second quadrant, the x-coordinate is negative. Thus, $\cos \alpha <0$.

Now, let us turn our attention to $\beta =3{\cos }^{-1}\left(\frac{4}{9}\right)$. Here is where the trap lies. Many students treat ${\cos }^{-1}$ exactly like ${\sin }^{-1}$, but they forget that ${\cos }^{-1}(x)$ is a strictly decreasing function.

Let us compare $\frac{4}{9}$ to $\frac{1}{2}$. Since $\frac{4}{9}<\frac{4.5}{9}=\frac{1}{2}$, the input is smaller. Because the function is decreasing, the output angle must be larger than the angle for $\frac{1}{2}$.

Multiplying by $3$, we get $\beta >3\times \frac{\pi }{3}=\pi$. We also know that $\frac{4}{9}$ is positive, so ${\cos }^{-1}\left(\frac{4}{9}\right)<\frac{\pi }{2}$.

Multiplying by $3$ gives $\beta <\frac{3\pi }{2}$. We have successfully trapped $\beta$ in the interval $\left(\pi ,\frac{3\pi }{2}\right)$.

This is the third quadrant! In the third quadrant, both sine and cosine are negative. This confirms that $\sin \beta <0$ is a correct statement, while $\cos \beta >0$ is false.

Summing the Angles

Finally, we must evaluate $\cos (\alpha +\beta )$. We have the ranges for both angles:

To find the range of the sum $\alpha +\beta$, we simply add the lower bounds and the upper bounds.

Lower bound: $\frac{\pi }{2}+\pi =\frac{3\pi }{2}$. Upper bound: $\frac{3\pi }{4}+\frac{3\pi }{2}=\frac{3\pi }{4}+\frac{6\pi }{4}=\frac{9\pi }{4}$.

So, $\alpha +\beta \in \left(\frac{3\pi }{2},\frac{9\pi }{4}\right)$.

Let us visualize this on the unit circle. $\frac{3\pi }{2}$ is the bottom of the circle (the negative y-axis). $\frac{9\pi }{4}$ is equivalent to $2\pi +\frac{\pi }{4}$, which is the first quadrant.

In the interval $\left(\frac{3\pi }{2},2\pi \right)$, we are in the fourth quadrant, where cosine is positive. In the interval $\left(2\pi ,2\pi +\frac{\pi }{4}\right)$, we are in the first quadrant, where cosine is also positive.

Since the entire range of $\alpha +\beta$ lies within regions where the x-coordinate is positive, we can conclude with absolute certainty that $\cos (\alpha +\beta )>0$.