MCQ
If $A, B$ and $C$ are interior angles of a triangle $\text{ABC,}$ then $\sin\Big(\frac{\text{B}+\text{C}}{2}\Big)=$
  • A
    $\sin\frac{\text{A}}{2}$
  • $\cos\frac{\text{A}}{2}$
  • C
    $-\sin\frac{\text{A}}{2}$
  • D
    $-\cos\frac{\text{A}}{2}$

Answer

Correct option: B.
$\cos\frac{\text{A}}{2}$
We know tht in triangle $\text{ABC}$
$\text{A+B+C}=180^\circ$
$\Rightarrow\text{B+C}=180^\circ-\text{A}$
$\Rightarrow\frac{\text{B+C}}{2}=\frac{90^\circ}{2}-\frac{\text{A}}{2}$
$\Rightarrow\sin\Big(\frac{\text{B+C}}{2}\Big)=\sin\Big(90^\circ-\frac{\text{A}}{2}\Big)$
Since $\sin(90^\circ-\text{A})=\cos{\text{A}}$
So,
$\Rightarrow\sin\Big(\frac{\text{B+C}}{2}\Big)=\cos\frac{\text{A}}{2}$
Hence the correct option is $(b)$

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