If $f\left(x\right)=\left\{\begin{array}{ccc}-2\sin x& \text{ if }& x\le -\frac{\pi }{2}\\ A\sin x+B& \text{ if }& -\frac{\pi }{2}<x<\frac{\pi }{2}\\ \cos x& \text{ if }& x\ge \frac{\pi }{2}\end{array}\right.$

is a continuous function then

$f\left(-\frac{\pi }{2}\right)=-2\sin \left(-\frac{\pi }{2}\right)=2$

$\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }(A\sin x+B)=-A+B$

so  $2=-A+B\dots \text{ (i) }$

Also $f\left(\frac{\pi }{2}\right)=\cos \frac{\pi }{2}=0$

$\underset{x\to \frac{\pi }{2}-}{\lim }f\left(x\right)=\underset{x\to \pi /2-}{\lim }(A\sin x+B)=A+B$

Hence A + B = 0            (ii)

Solving (i) and (ii), we get $A=-1, B=1$.