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Question 53
If $$(A + B + C) = 90$$°, then what is the value of $$\sin \left(\frac{A}{2}\right) \sin \left[\frac{(180 — B — C)}{2}\right] + \cos \left(\frac{A}{2}\right) \sin \frac{(B + C)}{2}$$?
Solution
We have :
A+B+C =90
Now (180-B-C)/2 = 90-(B-C)/2
Now sin (180-B-C)/2)= sin (90-(B+C)/2 )= cos (B+C)/2
So we get the expression as
$$\sin\ \frac{A}{2}\cos\ \frac{B+C}{2}+\cos\ \frac{A}{2}\sin\ \frac{B+C}{2}=\sin\ \left(\frac{A+B+C}{2}\right)\ =\sin\ 45\ =\ \frac{1}{\sqrt{\ 2}}$$
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