Browse Questions

# If $A+B+C=\pi$ then the value of $\sin\large\frac{A}{2}$$+\sin\large\frac{B}{2}$$+\sin\large\frac{C}{2}$ is equal to

$\begin{array}{1 1}(a)\;1+4\sin\big(\large\frac{B+C}{4}\big)\normalsize\sin\big(\large\frac{C+A}{4}\big)\normalsize \sin\big(\large\frac{A+B}{4}\big)\\(b)\;\;1+4\cos\big(\large\frac{B+C}{4}\big)\normalsize\cos\big(\large\frac{C+A}{4}\big)\normalsize \cos\big(\large\frac{A+B}{4}\big)\\(c)\;1-4\sin\big(\large\frac{B+C}{4}\big)\normalsize\sin\big(\large\frac{C+A}{4}\big)\normalsize \sin\big(\large\frac{A+B}{4}\big)\\(d)\;None\;of\;the\;above\end{array}$

Step 1:
We have
$\sin \large\frac{A}{2}+$$\sin\large\frac{B}{2}$$+\sin\large\frac{C}{2}$
$\Rightarrow (\sin\large\frac{A}{2}$$+\sin\large\frac{B}{2})$$+\sin\large\frac{C}{2}$
$\Rightarrow 2\sin\big(\large\frac{A+B}{4}\big)$$\cos\big(\large\frac{A-B}{4}$$+\cos\big(\large\frac{\pi}{2}-\frac{C}{2}\big)$
$\Rightarrow 2\sin\big(\large\frac{\pi-C}{4}\big)$$\cos\big(\large\frac{A-B}{4}$$+\cos\big(\large\frac{\pi-C}{2}\big)$
$\Rightarrow 2\sin\big(\large\frac{\pi-C}{4}\big)$$\cos\big(\large\frac{A-B}{4}$$+1-2\sin^2\big(\large\frac{\pi-C}{4}\big)$
$\Rightarrow 2\sin\big(\large\frac{\pi-C}{4}\big)$$\cos\big(\large\frac{A-B}{4}$$-2\sin^2\big(\large\frac{\pi-C}{4}\big)$$+1 \Rightarrow 2\sin\big(\large\frac{\pi-C}{4}\big)$$\bigg[\cos\big(\large\frac{A-B}{4}\big)$$-\sin \big(\large\frac{\pi-C}{4}\big)\bigg]$$+1$
$\Rightarrow 2\sin\big(\large\frac{\pi-C}{4}\big)$$\bigg[\cos\big(\large\frac{A-B}{4}\big)$$-\cos\{\large\frac{\pi}{2}- \big(\large\frac{\pi-C}{4}\big)\}\bigg]$$+1 \Rightarrow 2\sin\big(\large\frac{\pi-C}{4}\big)$$\bigg[\cos\big(\large\frac{A-B}{4}\big)$$-\cos\{\large\frac{\pi}{4}+ \large\frac{C}{4}\}\bigg]$$+1$
Step 2:
The above equation can be written as
$\Rightarrow 2\sin\big(\large\frac{\pi-C}{4}\big)$$\{2\sin\big(\large\frac{A-B+\pi+C}{8}\big)$$\sin\big(\large\frac{\pi+C-A+B}{8}\big)\}+$$1 \Rightarrow 2\sin\big(\large\frac{\pi-C}{4}\big)$$\big[2\sin\big(\large\frac{\pi-B}{4}\big)$$\sin\big(\large\frac{\pi-A}{4}\big)\big]$$+1$
$\Rightarrow 1+4\sin\big(\large\frac{\pi-A}{4}\big)$$\sin\big(\large\frac{\pi-B}{4}\big)$$\sin\big(\large\frac{\pi-C}{4}\big)$
$\Rightarrow 1+4\sin\big(\large\frac{B+C}{4}\big)\normalsize\sin\big(\large\frac{C+A}{4}\big)\normalsize \sin\big(\large\frac{A+B}{4}\big)$
Hence (a) is the correct answer.