In ABC prove that sinB-C2 - b-c2 cosA2-

ABC is a triangle. ∴ A+B+C=π ⇒sin Aa=sin Bb=sin Cc=k ⇒ 2cos(B+C2) sin(B C2)=(b c)k ⇒ 2sin(A2)sin(B C2)=(b c)k ⇒sin A=ak ⇒ 2si ...

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Angle Sum Property

In ABC prov...

Question

In $△ A B C$ prove that $sin (\frac{B - C}{2}) = (\frac{b - c}{2}) cos \frac{A}{2} .$

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△ A B C

sin (\frac{B - C}{2}) = (\frac{b - c}{a}) cos \frac{A}{2}

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tan \frac{A + B}{2} = - cot \frac{C}{2}

sin (\frac{B - C}{2}) = \frac{b - c}{a} cos \frac{A}{2}

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\frac{cos \frac{B + C}{2} cos \frac{C + A}{2} cos \frac{A + B}{2}}{sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}} = 1

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