Prove that $ \sin \left(C+\frac{A+B}{2}\right)=\sin \frac{A+B}{2} $

We know that,

$\sin(90^o+\theta)=\cos \theta$

$\sin(90^o-\theta)=\cos \theta$

In a triangle $A+B+C=180^o$

$\Rightarrow \frac{A+B+C}{2}=\frac{180^o}{2}=90^o$.....(i)

$\Rightarrow \frac{A+B}{2}=90^o-\frac{C}{2}$

Therefore,

LHS $=\sin (C+\frac{A+B}{2})$

$=\sin(\frac{A+B+2C}{2}$

$=\sin(\frac{A+B+C}{2}+\frac{C}{2})$

$=\sin(90^o+\frac{C}{2})$         [From (i)]

$=\cos \frac{C}{2}$

RHS $=\sin \frac{A+B}{2}$

$=\sin(90^o-\frac{C}{2})$       [From (ii)]

$=\cos \frac{C}{2}$

LHS $=$ RHS

Hence proved.