Question

If $A+B+C=180^{\circ}$, then $sin^{2} \frac{A}{2}+sin^{2} \frac{B}{2}+sin^{2} \frac{C}{2}=$

• A. $1-2\,cos \frac{A}{2}cos \frac{B}{2}cos \frac{C}{2}$
• B. $1-2\,sin \frac{A}{2}sin \frac{B}{2}sin\frac{C}{2}$
• C. $1-4\,sin \frac{A}{2}sin \frac{B}{2}sin\frac{C}{2}$
• D. $1-4\,cos \frac{A}{2}cos \frac{B}{2}cos \frac{C}{2}$

Answer 1

Answer: (B)

$L.H.S.=sin^{2} \frac{A}{2}+sin^{2} \frac{B}{2}+sin^{2} \frac{C}{2}$
$=\frac{1-cos\,A}{2}+\frac{1-cos\,B}{2}+\frac{1-cos\,C}{2}$
$=\frac{3-\left(cos\,A+cos\,B+cos\,C\right)}{2}$
$=\frac{3-S}{2}\quad\ldots\left(i\right)$
where $S=cosA + cosB + cosC $
$= \left(cosA + cosB\right) + cosC$
$=2\,cos\left(\frac{A+B}{2}\right)cos\left(\frac{A-B}{2}\right)+cos\left(2\cdot\frac{C}{2}\right)$
$=2\,cos\left(90^{\circ}-\frac{C}{2}\right)cos\left(\frac{A-B}{2}\right)+cos\left(2\cdot\frac{C}{2}\right)$
$=2\,sin \frac{C}{2}cos\left(\frac{A-B}{2}\right)+1-2\,sin^{2} \frac{C}{2}$
$=1+2\,sin \frac{C}{2}\left\{cos\left(\frac{A-B}{2}\right)-sin \frac{C}{2}\right\}$
$=1+2sin \frac{C}{2}\left\{cos\left(\frac{A-B}{2}\right)-sin\left(90^{\circ}-\frac{A+B}{2}\right)\right\}$
$=1+2sin \frac{C}{2}\left\{cos\left(\frac{A-B}{2}\right)-cos\left(\frac{A+B}{2}\right)\right\}$
$=1+2sin \frac{C}{2}\left\{-2\,sin \frac{A}{2}sin \left\{-\frac{B}{2}\right\}\right\}$
$=1+4sin \frac{A}{2}sin \frac{B}{2}sin \frac{C}{2}$
Then from $\left(i\right)$, we get
$L.H.S.=\frac{3-\left(1+4\,sin \frac{A}{2}sin \frac{B}{2}sin \frac{C}{2}\right)}{2}$
$=1-2sin \frac{A}{2}sin \frac{B}{2} sin \frac{C}{2}$