If $A$, $B$ and $C$ are interior angles of a $ riangle ABC$, then show that $\cos\left(\frac{B + C}{2} ight) = \sin\left(\frac{A}{2} ight)$.

Question

If $A$, $B$ and $C$ are interior angles of a $	riangle ABC$, then show that $\cos\left(\frac{B + C}{2}ight) = \sin\left(\frac{A}{2}ight)$.

Accepted Answer

Step 1: $A + B + C = 180° \Rightarrow B + C = 180° - A \Rightarrow \frac{B + C}{2} = 90° - \frac{A}{2}$
Step 2: $	herefore \cos\left(\frac{B + C}{2}ight) = \cos\left(90° - \frac{A}{2}ight) = \sin\left(\frac{A}{2}ight)$

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