proving $\cos (A+B)>0$, if given angles $A$ and $B$ If $\displaystyle A=3\sin^{-1}\left(\frac{6}{11}\right)$ and $\displaystyle B = 3\cos^{-1}\left(\frac{4}{9}\right),$ then proving $\cos (A+B)>0$
Attempt: $$ 3\sin^{-1}\left(\frac{1}{2}\right)<3\sin^{-1}\left(\frac{6}{11}\right)<3\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\Rightarrow \frac{\pi}{2}<A<\pi$$
same way $$\displaystyle 3\cos^{-1}\left(0\right)<3\cos^{-1}\left(\frac{4}{9}\right)<3\cos^{-1}\left(\frac{1}{2}\right)\Rightarrow \frac{\pi}{2}<B<\frac{3\pi}{2}$$
so $$\pi<A+B<\frac{5\pi}{2}$$
could some help me how to prove $\cos (A+B)>0,$ thanks

Let $\displaystyle \alpha=\sin^{-1}\left(\frac{6}{11}\right)=\frac{A}{3}$ and let $\displaystyle\beta=\cos^{-1}\left(\frac{4}{9}\right)=\frac{B}{3}$
Then since $\frac{1}{2}<\frac{6}{11}<\frac{\sqrt{2}}{2}$ we have $\frac{\pi}{6}<\alpha<\frac{\pi}{4}$.
And since $0<\frac{4}{9}<\frac{1}{2}$ we have $\frac{\pi}{3}<\beta<\frac{\pi}{2}$.
Addition of the two inequalities gives $\frac{\pi}{2}<\alpha+\beta<\frac{3\pi}{4}$.
Multiplication by $3$ gives $\frac{3\pi}{2}<3\alpha+3\beta<\frac{9\pi}{4}$.
Thus $\frac{3\pi}{2}<A+B<\frac{9\pi}{4}$ meaning that $A+B$ lies in either quadrant IV or quadrant I.
Thus $\cos(A+B)>0$.
What has been shown here is that, referring to the following diagram, for any standard angle $A$ such that the terminal side of $\dfrac{A}{3}$ lies in the blue region and for any standard angle $B$ such that the terminal side of $\dfrac{B}{3}$ lies in the green region it will be the case that $\cos(A+B)>0$. The "red" lines are the terminal sides of the particular $\dfrac{A}{3}$ and $\dfrac{B}{3}$ given in the question.

