Question
If A, B, C are the interior angles of a triangle ABC, prove that.
$\sin\Big(\frac{\text{B+C}}{2}\Big)=\cos\frac{\text{ A}}{2}$
$\sin\Big(\frac{\text{B+C}}{2}\Big)=\cos\frac{\text{ A}}{2}$
✓
Answer
$\sin\Big(\frac{\text{B+C}}{2}\Big)=\cos\Big(\frac{\text{ A}}{2}\Big)$
$\text{L.H.S}=\sin\Big(\frac{\text{B+C}}{2}\Big)$
$=\sin\Big(\frac{180^\circ-\text{A}}{2}\Big)$
$=\sin\Big(90^\circ-\frac{\text{A}}{2}\Big)$
$=\cos\frac{\text{A}}{2}$
$=\text{R.H.S}$
$\text{L.H.S}=\sin\Big(\frac{\text{B+C}}{2}\Big)$
$=\sin\Big(\frac{180^\circ-\text{A}}{2}\Big)$
$=\sin\Big(90^\circ-\frac{\text{A}}{2}\Big)$
$=\cos\frac{\text{A}}{2}$
$=\text{R.H.S}$
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