Question.
If $A, B$ and $C$ are interior angles of a triangle $A B C$, then show that $\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2} .$
If $A, B$ and $C$ are interior angles of a triangle $A B C$, then show that $\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2} .$
Solution:
A + B + C = 180°
$\Rightarrow \mathrm{B}+\mathrm{C}=180^{\circ}-\mathrm{A}$
$\Rightarrow \frac{\mathbf{B}+\mathbf{C}}{2}=\frac{180^{\circ}-\mathbf{A}}{2} \Rightarrow \frac{\mathbf{B}+\mathbf{C}}{2}=\left(\mathbf{M}^{\circ}-\frac{\mathbf{A}}{2}\right)$
$\Rightarrow \sin \left(\frac{\mathbf{B}+\mathbf{C}}{\mathbf{2}}\right)=\sin \left(\mathbf{8 0}^{\circ}-\frac{\mathbf{A}}{\mathbf{2}}\right)=\cos \frac{\mathbf{A}}{\mathbf{2}}$
$\left\{\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right\}$
A + B + C = 180°
$\Rightarrow \mathrm{B}+\mathrm{C}=180^{\circ}-\mathrm{A}$
$\Rightarrow \frac{\mathbf{B}+\mathbf{C}}{2}=\frac{180^{\circ}-\mathbf{A}}{2} \Rightarrow \frac{\mathbf{B}+\mathbf{C}}{2}=\left(\mathbf{M}^{\circ}-\frac{\mathbf{A}}{2}\right)$
$\Rightarrow \sin \left(\frac{\mathbf{B}+\mathbf{C}}{\mathbf{2}}\right)=\sin \left(\mathbf{8 0}^{\circ}-\frac{\mathbf{A}}{\mathbf{2}}\right)=\cos \frac{\mathbf{A}}{\mathbf{2}}$
$\left\{\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right\}$