Question

If A, B and C are interior angles of a triangle ABC, then $\sin \left(\frac{B+C}{2}\right)=$?

• A. $\cos \frac{B}{2}$.
• B. $cosA$.
• C. $\cos \frac{A}{2}$.
• D. $\sin \frac{A}{2}$.

Answer 1

Answer: (C)

$\cos \frac{A}{2}$.

Let in $△ABC$

Sum of angle of a triangle is =${180}^0$ (by angle sum property)

$A+B+C={180}^0$

$B+C={180}^0-A$

multiplying both side by $\frac{1}{2}$

$\frac{B+C}{2}=\frac{{180}^0-A}{2}$

$\frac{B+C}{2}=\frac{{180}^0}{2}-\frac{A}{2}$

$\frac{B+C}{2}={90}^0-\frac{A}{2}.........\left(1\right)$

put sin both side we have

$sin\frac{B+C}{2}=sin({90}^0-\frac{A}{2})$

Taking R.H.S

$sin({90}^0-\frac{A}{2})=cos\left(\frac{A}{2}\right)$ (by using property)

hence prove

$sin\left(\frac{B+C}{2}\right)=cos\left(\frac{A}{2}\right)$