If $\sin \left(\frac{2A + B}{2}\right) = \cos \left(\frac{2A - B}{2}\right) = \frac{\sqrt{3}}{2}, 0^\circ < \frac{2A + B}{2} < 90^\circ$ and $0^\circ < \frac{2A + B}{2} < 90^\circ$ then find the value of $\sin[3(A - B)]$.

$\frac{1}{\sqrt{2}}$

We are given that :-

sin( \(\frac{2A + B }{2}\) ) = cos( \(\frac{2A - B }{2}\) ) = \(\frac{ √3}{2}\)

{ we know, sin 60º = cos30º = \(\frac{ √3}{2}\) }

So, ( \(\frac{2A + B }{2}\) ) = 60º

2A + B = 120º   ----(1)

& ( \(\frac{2A - B }{2}\) ) = 30º 

2A - B = 60º    ----(2)

On adding equation 1 and 2 ,

4A = 120º + 60º

4A = 180º

A = 45º

Now , put value of A in eqn 1 .

2 × 45º  + B = 120º

B = 30º

Now,

sin [ 3 (A - B) ]

= sin [ 3 (45º  - 30º) ] 

= sin 45º

= \(\frac{1}{√2}\)