Question
If A, B, C are the interior angles of a $\triangle\text{ABC},$ show that:$\sin\frac{\text{B+C}}{2}=\cos\frac{\text{A}}{2}$
✓
Answer
We have to prove: $\sin\Big(\frac{\text{B+C}}{2}\Big)=\cos\frac{\text{A}}{2}$
Since we know that in triangle ABC
$\text{A + B + C}=180^\circ$
$\Rightarrow\ \text{B + C}=180^\circ-\text{A}$
Dividing by 2 on both sides, we get
$\Rightarrow\ \frac{\text{B+C}}{2}=90^\circ-\frac{\text{A}}{2}$
$\Rightarrow\ \sin \frac{\text{B+C}}{2}=\sin\Big(90^\circ-\frac{\text{A}}{2}\Big)$
$\Rightarrow\ \sin\frac{\text{B+C}}{2}=\cos\frac{\text{A}}{2}$
Since we know that in triangle ABC
$\text{A + B + C}=180^\circ$
$\Rightarrow\ \text{B + C}=180^\circ-\text{A}$
Dividing by 2 on both sides, we get
$\Rightarrow\ \frac{\text{B+C}}{2}=90^\circ-\frac{\text{A}}{2}$
$\Rightarrow\ \sin \frac{\text{B+C}}{2}=\sin\Big(90^\circ-\frac{\text{A}}{2}\Big)$
$\Rightarrow\ \sin\frac{\text{B+C}}{2}=\cos\frac{\text{A}}{2}$
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