In - A B C- prove that-a sin A 2 sin B-C 2-b sin B 2 sin C-A 2-c sin C 2 sin A-B 2-0

Consider asinA2sinB C2+bsinB2sinC A2+csinC2sinA B2 =ksinAsinA2sinB C2+sinBsinB2sinC A2+sinCsinC2sinA B2=ksin #960; B+CsinA2sinB C2+sin #960; C+A #160;sinB ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

In Δ A B C, p...

Question

In ∆ABC, prove that:

$a \sin \frac{A}{2} \sin (\frac{B - C}{2}) + b \sin \frac{B}{2} \sin (\frac{C - A}{2}) + c \sin \frac{C}{2} \sin (\frac{A - B}{2}) = 0$ .

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Solution

Consider
$a \sin \frac{A}{2} \sin (\frac{B - C}{2}) + b \sin \frac{B}{2} \sin (\frac{C - A}{2}) + c \sin \frac{C}{2} \sin (\frac{A - B}{2})$
$= k [\sin A \sin \frac{A}{2} \sin (\frac{B - C}{2}) + \sin B \sin \frac{B}{2} \sin (\frac{C - A}{2}) + \sin C \sin \frac{C}{2} \sin (\frac{A - B}{2})] = k [\sin \{π - (B + C)\} \sin \frac{A}{2} \sin (\frac{B - C}{2}) + \sin \{π - (C + A)\} \sin \frac{B}{2} \sin (\frac{C - A}{2}) + \sin \{π - (A + B)\} \sin \frac{C}{2} \sin (\frac{A - B}{2})] (∵ A + B + C = π) = k [\sin (B + C) \sin \frac{A}{2} \sin (\frac{B - C}{2}) + \sin (A + C) \sin \frac{B}{2} \sin (\frac{C - A}{2}) + \sin (A + B) \sin \frac{C}{2} \sin (\frac{A - B}{2})] = k [2 \sin (\frac{B + C}{2}) \cos (\frac{B - C}{2}) \sin \frac{A}{2} \sin (\frac{B - C}{2}) + 2 \sin (\frac{A + C}{2}) \cos (\frac{C - A}{2}) \sin \frac{B}{2} \sin (\frac{C - A}{2}) + 2 \sin (\frac{A + B}{2}) \cos (\frac{A - B}{2}) \sin \frac{C}{2} \sin (\frac{A - B}{2})] = 2 k [\sin (\frac{B + C}{2}) \sin \frac{A}{2} \sin \frac{A}{2} \sin (\frac{B - C}{2}) + \sin (\frac{A + C}{2}) \sin \frac{B}{2} \sin \frac{B}{2} \sin (\frac{C - A}{2}) + \sin (\frac{A + B}{2}) \sin \frac{C}{2} \sin \frac{C}{2} \sin (\frac{A - B}{2})] = 2 k [\sin (\frac{B + C}{2}) \sin (\frac{B - C}{2}) \sin^{2} \frac{A}{2} + \sin (\frac{A + C}{2}) \sin (\frac{C - A}{2}) \sin^{2} \frac{B}{2} + \sin (\frac{A + B}{2}) \sin (\frac{A - B}{2}) \sin^{2} \frac{C}{2}] = 2 k \sin^{2} \frac{A}{2} (\sin^{2} \frac{B}{2} - \sin^{2} \frac{C}{2}) + 2 k \sin^{2} \frac{B}{2} (\sin^{2} \frac{C}{2} - \sin^{2} \frac{A}{2}) + 2 k \sin^{2} \frac{C}{2} (\sin^{2} \frac{A}{2} - \sin^{2} \frac{B}{2}) = 2 k (\sin^{2} \frac{A}{2} \sin^{2} \frac{B}{2} - \sin^{2} \frac{A}{2} \sin^{2} \frac{C}{2} + \sin^{2} \frac{B}{2} \sin^{2} \frac{C}{2} - \sin^{2} \frac{A}{2} \sin^{2} \frac{B}{2} + \sin^{2} \frac{A}{2} \sin^{2} \frac{C}{2} - \sin^{2} \frac{C}{2} \sin^{2} \frac{B}{2}) = k (0) = 0$

Hence proved.

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Similar questions

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In ∆ABC, prove that: a sinA2 sin B-C2+b sin B2 sin C-A2+c sin C2 sin A-B2=0 .

In ∆ABC, prove that:

$a \sin \frac{A}{2} \sin (\frac{B - C}{2}) + b \sin \frac{B}{2} \sin (\frac{C - A}{2}) + c \sin \frac{C}{2} \sin (\frac{A - B}{2}) = 0$ .