In any - ABC- prove that -a sinA-2sinB-C-2-b sinB-2sinC-A-2- c sinC-2sinA-B-2-0-

We know. a sin B=b sin A, c sin B=b sin C, a sin C=c sin B a sinA/2sin(B C/2 )+b sin B/2sin(C A/2 )+ c sin C/2 sin(A B/2 )=0 LHS =a sin(π (B+C )/2 )sin(B C ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Sine Rule

In any Δ ABC,...

Question

$In any Δ A B C, prove that : a s i n \frac{A}{2} s i n (\frac{B - C}{2}) + b s i n \frac{B}{2} s i n (\frac{C - A}{2}) + c s i n \frac{C}{2} s i n (\frac{A - B}{2}) = 0.$

Open in App

Solution

$We know . a s i n B = b s i n A, c s i n B = b s i n C, a s i n C = c s i n B a s i n \frac{A}{2} s i n (\frac{B - C}{2}) + b s i n \frac{B}{2} s i n (\frac{C - A}{2}) + c s i n \frac{C}{2} s i n (\frac{A - B}{2}) = 0 L H S = a s i n (\frac{π - (B + C)}{2}) s i n (\frac{B - C}{2}) + b s i n (\frac{π - (C + A)}{2}) s i n (\frac{C - A}{2}) + c s i n (\frac{π - (A + B)}{2}) s i n (\frac{A - B}{2}) = a c o s (\frac{B + C}{2}) s i n (\frac{B - C}{2}) + b c o s (\frac{C + A}{2}) s i n (\frac{C - A}{2}) + c c o s (\frac{A + B}{2}) s i n (\frac{A - B}{2}) = a (s i n B - s i n C) + b (s i n C - s i n A) + c (s i n A - s i n B) = a s i n B - a s i n C + b s i n C - b s i n A + c s i n A - c s i n B = b s i n A - a s i n C + b s i n C - b s i n A + a s i n C - b s i n C = 0 = R H S .$

Suggest Corrections

Similar questions

Q. For any three angles A, B, C prove that

sin A + sin B + sin C - sin (A + B + C) = 4 sin {(A + B) / 2} sin {(B + C) / 2} sin {(C + A) / 2}

Prove that ${sin}^{2} (\frac{A}{2}) + {sin}^{2} (\frac{B}{2}) + {sin}^{2} (\frac{C}{2}) = 1 - 2 sin (\frac{A}{2}) sin \frac{B}{2}) sin (\frac{C}{2})$ .