If the function
\(f(x)= -2 \sin x \text {, if } x \leq \frac{-\pi}{2} \ A \sin x+B , \text { if } \frac{-\pi}{2} \lt x\) \(\lt \frac{\pi}{2} \ \cos x , \text { if } x \geq \frac{\pi}{2}\) is continuous everywhere, then the values of A and B are respectively

Since \(\mathrm{f}(x)\) is continuous everywhere.
\(\therefore \quad \mathrm{f}(x)\) is continuous at \(x=-\frac{\pi}{2}\) and \(x=\frac{\pi}{2}\).
\(\begin{aligned}
\therefore \quad & \lim _{x \rightarrow \frac{-\pi^{-}}{2}} \mathrm{f}(x)=\lim _{x \rightarrow \frac{-\pi^{+}}{2}} \mathrm{f}(x) \\
& \Rightarrow \lim _{x \rightarrow \frac{-\pi}{2}}(-2 \sin x)=\lim _{x \rightarrow \frac{-\pi}{2}}(\mathrm{~A} \sin x+\mathrm{B}) \\
\Rightarrow & -2(-1)=\mathrm{A}(-1)+\mathrm{B} \\
& \Rightarrow-\mathrm{A}+\mathrm{B}=2...(i)
\end{aligned}\)
\(\begin{aligned}
& \text { Also, } \lim _{x \rightarrow \frac{\pi^{-}}{2}} \mathrm{f}(x)=\lim _{x \rightarrow \frac{\pi^{+}}{2}} \mathrm{f}(x) \\
& \Rightarrow \lim _{x \rightarrow \frac{\pi}{2}}(\mathrm{~A} \sin x+\mathrm{B})=\lim _{x \rightarrow \frac{\pi}{2}}(\cos x) \\
& \Rightarrow \mathrm{A}(1)+\mathrm{B}=0 ...(ii)\\
& \Rightarrow \mathrm{~A}+\mathrm{B}=0
\end{aligned}\)
From (i) and (ii), we get
\(\mathrm{A}=-1, \mathrm{~B}=1\)